Answer: $f'(x) = 0$

Definition of critical point: A critical point of a differentiable function occurs where the first derivative equals zero: $f'(c) = 0$. Why the correct answer works: At a maximum or minimum on an open interval, the tangent line is horizontal, so $f'(x) = 0$. Why distractors fail: Option A sets the function itself to zero, which is a root, not a critical point. Option C sets the second derivative to zero, which identifies a possible inflection point. Option D would mean the function is increasing, not at an extremum.

Answer: $T(x) = \dfrac{\sqrt{9 + x^2}}{3} + \dfrac{8 - x}{5}$

Model the rowing distance: The rowing distance from the boat (3 km offshore) to a point $x$ km down the coast is $\sqrt{3^2 + x^2} = \sqrt{9 + x^2}$ by the Pythagorean theorem. Write the time function: Time = distance/speed. Rowing time $= \frac{\sqrt{9 + x^2}}{3}$, walking time $= \frac{8 - x}{5}$. Total: $T(x) = \frac{\sqrt{9 + x^2}}{3} + \frac{8 - x}{5}$. Why distractors fail: Option B swaps the speeds. Option C uses $3 + x$ instead of $\sqrt{9 + x^2}$, adding distances instead of using the Pythagorean theorem. Option D uses $x$ as the rowing distance, ignoring the offshore distance.

Answer: The lifeguard should run closer to the point on shore nearest the swimmer before entering the water at a relatively steep angle

Interpret the equation: Since running speed (8 m/s) is much greater than swimming speed (2 m/s), $\sin\theta_1$ must be much larger than $\sin\theta_2$. This means $\theta_1$ is large (shallow angle to the shore normal) while $\theta_2$ is small (steep entry into water). Why the correct answer works: The lifeguard covers more distance on land (where they are fast) by running closer to the point nearest the swimmer, then enters the water steeply to minimize the slower swimming distance. Why distractors fail: Option A is suboptimal — entering perpendicular means $\theta_2 = 0$, which isn't generally optimal. Option B ignores the speed difference. Option D is false — the angle greatly affects time.

Answer: $x = 40$

Form the profit function: $P(x) = R(x) - C(x) = (50x - 0.5x^2) - (10x + 100) = -0.5x^2 + 40x - 100$. Differentiate and solve: $P'(x) = -x + 40 = 0 \implies x = 40$. Verify maximum: $P''(x) = -1 < 0$, confirming a maximum at $x = 40$. Why distractors fail: Option A ($x = 50$) maximizes revenue, not profit. Option B ($x = 80$) is where revenue equals zero. Option D ($x = 20$) is half the correct answer, a common arithmetic slip.

Answer: Apply single-variable differentiation to find critical points

Purpose of reducing variables: By substituting the constraint into the objective function, you convert a two-variable problem into one that depends on a single variable, enabling the use of standard Calculus 1 techniques. Why the correct answer works: Once the objective is $f(x)$ in one variable, you can compute $f'(x)$, set it to zero, and solve for critical points. Why distractors fail: Option A is algebraic, not calculus-based. Option C is wrong — the constraint is used in the reduction step, not discarded. Option D describes multivariable graphing, which is the opposite goal.

Answer: No, the constraint relates the variables but is not the function being optimized

Role of the constraint vs. objective: The constraint is used to eliminate a variable; the objective function is what gets differentiated to find extrema. Why the correct answer works: Differentiating the constraint gives the relationship between the rates of change of the variables, not the critical points of the quantity being optimized. Why distractors fail: Options A and B both claim the approach is valid, which it is not in the standard single-variable method. Option D's reasoning is incomplete — while the constraint often equals a constant, the real issue is that differentiating it does not optimize the objective.

Answer: $\text{Cost}(r) = 2c\pi r^2 + \dfrac{2c}{r}$

Identify constraint and surfaces: Volume: $\pi r^2 h = 1 \implies h = \frac{1}{\pi r^2}$. The tank is open-top, so surfaces are one base ($\pi r^2$) and the lateral side ($2\pi rh$). Write the cost function: Cost $= 2c(\pi r^2) + c(2\pi rh) = 2c\pi r^2 + 2c\pi r \cdot \frac{1}{\pi r^2} = 2c\pi r^2 + \frac{2c}{r}$. Why distractors fail: Option B uses $c$ instead of $2c$ for the base. Option C uses $c$ instead of $2c \cdot \pi r / (\pi r^2)$ for the side term — it should be $2c/r$. Option D is not reduced to one variable since $h$ still appears.

Answer: Yes, the cost function correctly weights the top and bottom at triple the side cost

Verify the cost function: Top and bottom area: $2\pi r^2$, at triple cost: $3 \cdot 2\pi r^2 = 6\pi r^2$. Side area: $2\pi rh$, at base cost: $1 \cdot 2\pi rh$. Total: $6\pi r^2 + 2\pi rh$. ✓ Verify the reduction: Constraint: $\pi r^2 h = 1000 \implies h = \frac{1000}{\pi r^2}$. Substituting: $K(r) = 6\pi r^2 + 2\pi r \cdot \frac{1000}{\pi r^2} = 6\pi r^2 + \frac{2000}{r}$. ✓ Why distractors fail: Option B incorrectly claims only one circle is tripled — both top and bottom use the expensive material. Option C applies the triple cost to the wrong surface. Option D is wrong because 1 liter $= 1000$ cm³.

Answer: Checking the sign of the second derivative at the critical point

Verification methods: The second derivative test is a standard way to classify critical points: if $f''(c) > 0$, the point is a local minimum; if $f''(c) < 0$, it is a local maximum. Why the correct answer works: The sign of $f''$ at the critical point directly tells you the concavity there, confirming the nature of the extremum. Why distractors fail: Option A checks feasibility but does not classify the extremum. Option C merely confirms it is a critical point, not whether it is a max or min. Option D applies to closed-interval problems via the Candidates Test but does not classify the critical point itself.

Answer: $S = 2\pi r^2 + 2\pi rh$

Surface area of a closed cylinder: A closed cylinder has two circular ends (each $\pi r^2$) and a lateral surface ($2\pi rh$), so $S = 2\pi r^2 + 2\pi rh$. Why the correct answer works: Since the can is closed, both top and bottom contribute $\pi r^2$ each, giving $2\pi r^2 + 2\pi rh$. Why distractors fail: Option A only counts one circular end. Option C only counts the lateral surface. Option D is the volume formula, not the surface area.

Answer: $P(x) = x(50 - x)$

Set up the problem: Let the two numbers be $x$ and $y$ with $x + y = 50$. Then $y = 50 - x$ and the product is $P = xy = x(50 - x)$. Why the correct answer works: $P(x) = x(50 - x) = 50x - x^2$ correctly represents the product of two positive numbers summing to 50. Why distractors fail: Option A gives $x(x - 50)$, which reverses the sign and yields negative values for $0 < x < 50$. Option C is the sum, not the product. Option D is the square of only one of the numbers.

Answer: Because the objective function often approaches unfavorable values at the endpoints of the feasible domain, so a single interior critical point must be the global extremum

Behavior at domain boundaries: In many applied problems, as the variable approaches the endpoints of the feasible interval, the objective function tends to zero, infinity, or another extreme unfavorable value. Why the correct answer works: If the function goes to unfavorable values at both ends and there is only one interior critical point, that point must be the global maximum or minimum on the interval. Why distractors fail: Option A is too restrictive — the second derivative is not always negative. Option B is false — the objective is typically nonlinear. Option D confuses the constraint with the derivative's properties.

Answer: $S(x) = 2x^2 + \dfrac{4000}{x}$

Identify the constraint: Volume: $x^2 h = 1000 \implies h = \frac{1000}{x^2}$. Write surface area in one variable: $S = 2x^2 + 4xh = 2x^2 + 4x \cdot \frac{1000}{x^2} = 2x^2 + \frac{4000}{x}$. Why distractors fail: Option B only counts one base ($x^2$ instead of $2x^2$). Option C uses $2xh$ instead of $4xh$ (two sides instead of four). Option D is the surface area of a cube, ignoring the height constraint.

Answer: It ensures the solution is physically or contextually meaningful

Context of domain: In applied optimization, variables like length, width, or quantity must be non-negative and may have upper bounds set by the constraint. The domain restricts the variable to physically meaningful values. Why the correct answer works: For example, if $w$ is a width, then $w > 0$ and $w < 100$ in a fencing problem. A critical point outside this interval would be meaningless. Why distractors fail: Option A is incorrect — the derivative formula depends on the function, not the domain. Option C is false — the domain does not control the number of critical points. Option D is wrong — the second derivative test is still needed to classify critical points.

Answer: Objective function

Define the key term: In optimization, the function whose maximum or minimum value we seek is called the objective function. Why the correct answer works: The objective function represents the quantity to be optimized — for example, area, cost, or distance. Why distractors fail: Option A (constraint equation) is the relation that limits the variables, not the function being optimized. Option C (critical function) is not a standard term. Option D (derivative function) is the tool used to find extrema, not the function being optimized.

Answer: Use the constraint to eliminate one variable from the objective function

Standard optimization workflow: Once you have an objective function in two variables and a constraint, you substitute the constraint into the objective to reduce it to a single-variable function. Why the correct answer works: Reducing to one variable allows you to use single-variable calculus (take one derivative, set it to zero) to find critical points. Why distractors fail: Option A has no mathematical basis. Option B may help with visualization but is not the standard procedural step. Option D describes a Lagrange multiplier–like approach, which is beyond the scope of Calculus 1 optimization.

Answer: The shortest distance does not always correspond to the shortest time when speed varies along the path

Time vs. distance optimization: Time $= \int \frac{ds}{v}$, where $v$ varies along the path. Minimizing this integral may favor a longer path through a faster region over a shorter path through a slower region. Why the correct answer works: When speed depends on position, the relationship between distance and time is nonlinear. The path of minimum time may curve to spend more distance in the faster region. Why distractors fail: Option B is false — the derivative need not be zero along the straight line. Option C mischaracterizes the issue — the constraint doesn't change; the objective function structure does. Option D is false — the second derivative test works normally.

Answer: No, because maximum revenue occurs at $x = 100$ while maximum profit occurs at $x = 75$

Find max revenue: $R'(x) = 200 - 2x = 0 \implies x = 100$. Find max profit: $P(x) = R(x) - C(x) = -x^2 + 150x - 1000$. $P'(x) = -2x + 150 = 0 \implies x = 75$. Evaluate the claim: Since maximum revenue occurs at $x = 100$ and maximum profit at $x = 75$, they do not coincide. The claim is incorrect. Why distractors fail: Options A and B both affirm the claim, which is false. Option D is wrong because $R(x) = 200x - x^2$ does have a maximum at $x = 100$.

Answer: Snell's law of refraction

Connection to physics: Minimizing travel time when crossing between two media with different speeds is mathematically identical to Snell's law, which describes how light bends when passing between media with different refractive indices. Why the correct answer works: Snell's law states $\frac{\sin\theta_1}{v_1} = \frac{\sin\theta_2}{v_2}$, the same condition that arises from minimizing $T(x)$ in the road-field problem. Why distractors fail: Option A ($F = ma$) relates to force and acceleration, not path optimization. Option B (conservation of energy) is unrelated to path selection. Option D (superposition) applies to wave or force combinations.

Answer: A square with side length $P/4$

Set up and solve: Let width $= w$, length $= l = P/2 - w$. Area $A = w(P/2 - w)$. $A'(w) = P/2 - 2w = 0 \implies w = P/4$. Then $l = P/4$. Why the correct answer works: The optimal rectangle is a square, regardless of the value of $P$. This is a classic result: for fixed perimeter, the square maximizes area. Why distractors fail: Option A describes a degenerate rectangle with minimal area. Option C gives area $= (P/3)(P/6) = P^2/18$, less than the square's area $P^2/16$. Option D is wrong — the square is always optimal.

Answer: It limits the possible values of the variables involved

Understand the constraint: A constraint equation expresses a relationship between the variables that must be satisfied, thereby restricting the domain of possible solutions. Why the correct answer works: The constraint limits the feasible values — for instance, a fixed perimeter limits the dimensions of a rectangle. Why distractors fail: Option A confuses the constraint with the objective function. Option C describes the result of solving the problem, not the role of the constraint. Option D refers to the second derivative test, not the constraint.

Answer: $r^2 = 2Rh - h^2$

Geometric relationship: Place the sphere centered at the origin. The cone's apex is at the top of the sphere at $(0, R)$ and its base is at height $y = R - h$. A point on the base circle satisfies $r^2 + (R - h)^2 = R^2$. Derive the constraint: $r^2 = R^2 - (R - h)^2 = R^2 - R^2 + 2Rh - h^2 = 2Rh - h^2$. Why distractors fail: Option A treats the slant height and height as the hypotenuse $R$, which would describe a cone inscribed differently. Option C is dimensionally inconsistent ($r^2$ vs. $h$). Option D introduces an angle not given in the problem.

Answer: The student should also consider $x = -2$ as a solution to $3x^2 = 12$

Solve the equation completely: $3x^2 - 12 = 0 \implies x^2 = 4 \implies x = \pm 2$. Why the correct answer works: The student found only $x = 2$ but missed $x = -2$. Even if $x = -2$ may later be excluded by domain constraints, it should be identified first. Why distractors fail: Option A overlooks the missing solution. Option C cannot be determined without seeing the original function (the derivative could be correct). Option D is irrelevant — the quotient rule is not needed here.

Answer: $A = 100w - w^2$

Set up the constraint: From $2l + 2w = 200$, solve for $l$: $l = 100 - w$. Express area in one variable: Area $A = l \cdot w = (100 - w)w = 100w - w^2$. Why distractors fail: Option A uses $200w$ instead of $100w$, likely from dividing the perimeter incorrectly. Option C has the signs reversed. Option D incorrectly doubles the $w^2$ term, suggesting a perimeter of $l + 2w = 200$.

Answer: $h/r = 2$

Compute the ratio: $\frac{h}{r} = \frac{\left(\frac{4V}{\pi}\right)^{1/3}}{\left(\frac{V}{2\pi}\right)^{1/3}} = \left(\frac{4V/\pi}{V/(2\pi)}\right)^{1/3} = \left(\frac{4V}{\pi} \cdot \frac{2\pi}{V}\right)^{1/3} = (8)^{1/3} = 2$. Geometric interpretation: The optimal closed cylinder has height equal to its diameter ($h = 2r$). This is a well-known result in optimization. Why distractors fail: Option A ($h = r$) would mean the height equals the radius, which is not optimal. Option C introduces $\pi$ with no justification. Option D gives $h = 4r$, which is too tall.

Answer: $70$ dollars

Write revenue and profit: $R(x) = px = (120 - 2x)x = 120x - 2x^2$. Profit: $P(x) = 120x - 2x^2 - 20x - 200 = -2x^2 + 100x - 200$. Find optimal quantity: $P'(x) = -4x + 100 = 0 \implies x = 25$. Find optimal price: $p = 120 - 2(25) = 70$ dollars. Why distractors fail: Option B ($60$) corresponds to $x = 30$. Option C ($50$) corresponds to $x = 35$. Option D ($25$) confuses the quantity with the price.

Answer: $x^2 + 4xh = 48$

Identify all surfaces: The box has a square base of area $x^2$ and four rectangular sides each of area $xh$. There is no top. Write the constraint: Total surface area $= x^2 + 4xh = 48$. Why distractors fail: Option B includes $2x^2$, which would count both a top and bottom. Option C uses $2xh$ instead of $4xh$, counting only two sides. Option D uses $xh$, counting only one side.

Answer: $(1, 1)$

Set up the distance (squared) function: Minimize $D^2 = (x - 3)^2 + (x^2)^2 = (x - 3)^2 + x^4$. Minimizing $D^2$ is equivalent to minimizing $D$. Differentiate and solve: $\frac{d}{dx}D^2 = 2(x - 3) + 4x^3 = 4x^3 + 2x - 6 = 0$, i.e., $2x^3 + x - 3 = 0$. Testing $x = 1$: $2 + 1 - 3 = 0$ ✓. So $x = 1$, $y = 1$. Verify: Factor: $2x^3 + x - 3 = (x - 1)(2x^2 + 2x + 3) = 0$. The quadratic $2x^2 + 2x + 3$ has discriminant $4 - 24 < 0$, so $x = 1$ is the only real critical point. The second derivative is $12x^2 + 2 > 0$, confirming a minimum. Why distractors fail: Option B: distance from $(2, 4)$ to $(3, 0)$ is $\sqrt{1 + 16} = \sqrt{17} \approx 4.12$. Option C: distance from $(1.5, 2.25)$ to $(3, 0)$ is $\sqrt{2.25 + 5.0625} \approx 2.70$. Option D: distance from $(0, 0)$ to $(3, 0)$ is $3$. The minimum distance at $(1, 1)$ is $\sqrt{4 + 1} = \sqrt{5} \approx 2.24$.

Answer: Two

Identify the decisions: The pipeline meets the highway at some point and leaves the highway at another point. Let the entry point be $x_1$ km east and the exit point be $x_2$ km east. These are the two decision variables. Why the correct answer works: The total cost depends on where the pipeline joins the highway ($x_1$) and where it leaves ($x_2$). These two variables fully determine the three-segment path. Why distractors fail: Option A undercounts — one variable would fix only one junction point. Option C would be correct only if there were an additional route decision. Option D overcounts.

Answer: The objective function will have two or more independent variables, making single-variable calculus inapplicable

Role of the constraint: The constraint is needed to eliminate one variable. Without it, the objective function remains a function of two or more variables. Why the correct answer works: Differentiating a two-variable function with respect to one variable while the other is also unknown and unconstrained does not yield meaningful critical points for the optimization. Why distractors fail: Option A — the derivative may still exist as a partial derivative but the approach fails. Option C — skipping the constraint does not merely lengthen the process; it changes the problem. Option D — the issue is more fundamental than extraneous solutions.

Answer: Apply the first derivative test to classify $x = 4$

Second derivative test is inconclusive: When $f''(c) = 0$, the second derivative test gives no information about whether $c$ is a max, min, or neither. Why the correct answer works: The first derivative test checks the sign of $f'(x)$ on either side of $x = 4$. If $f'$ changes from positive to negative, $x = 4$ is a local max; negative to positive, a local min. Why distractors fail: Option A is premature — $f''(c) = 0$ does not mean 'no extremum,' it means the test is inconclusive. Option C uses $x = -2$, which is outside the feasible domain. Option D skips the classification entirely.

Answer: The critical point is extraneous and lies outside the feasible domain

Domain restrictions in applied problems: Physical dimensions must be positive. A critical point $x = -5$ violates this requirement. Why the correct answer works: The critical point is outside the feasible domain (e.g., $x > 0$), so it must be discarded. The student should check for errors or look for endpoints or other critical points in the valid domain. Why distractors fail: Option A is nonsensical — dimensions cannot be negative. Option C is wrong — the second derivative test cannot make an infeasible point feasible. Option D is premature — there may still be a solution at a domain endpoint or the setup may need revisiting.

Answer: $V(x) = x(20 - 2x)(30 - 2x)$

Visualize the construction: Cutting squares of side $x$ from each corner removes $x$ from both ends of each dimension. The resulting base is $(20 - 2x)$ by $(30 - 2x)$, and the height is $x$. Write the volume: $V(x) = x(20 - 2x)(30 - 2x)$ with domain $0 < x < 10$. Why distractors fail: Option A subtracts $x$ instead of $2x$ from each dimension. Option C has an extra factor of 2. Option D factors out 2 from each dimension but then omits those factors.

Answer: Width $= 14$ cm, Height $= 21$ cm

Define variables: Let the printed area have width $w$ and height $h$. Then $wh = 150$. The poster has total width $w + 4$ and total height $h + 6$. Write the objective: $A = (w + 4)(h + 6)$. Using $h = 150/w$: $A(w) = (w + 4)(150/w + 6) = 150 + 6w + 600/w + 24$. Minimize: $A'(w) = 6 - 600/w^2 = 0 \implies w^2 = 100 \implies w = 10$. Then $h = 15$. Total: width $= 10 + 4 = 14$, height $= 15 + 6 = 21$. Why distractors fail: Option B uses the print width (10) as the poster width. Option C uses the print height (15) as the poster height. Option D does not correspond to a critical point of the area function.

Answer: $A(\theta) = 2\cos\theta + \sin\theta\cos\theta$

Set up the trapezoid dimensions: With angle $\theta$ from the vertical, each side contributes horizontal extension $\sin\theta$ and height $\cos\theta$. The top width $= 2 + 2\sin\theta$ and the height $= \cos\theta$. Compute the area: Trapezoid area $= \frac{1}{2}(b_1 + b_2) \cdot h = \frac{1}{2}(2 + 2 + 2\sin\theta)\cos\theta = \frac{1}{2}(4 + 2\sin\theta)\cos\theta = (2 + \sin\theta)\cos\theta = 2\cos\theta + \sin\theta\cos\theta$. Why distractors fail: Option B incorrectly uses $\sin\theta$ for height. Option C uses $(2 + \sin\theta)\sin\theta$, confusing height and width components. Option D uses $\sin^2\theta$ instead of $\sin\theta\cos\theta$.

Answer: $A(x) = \frac{8x}{3}\sqrt{9 - x^2}$

Express $y$ using the ellipse equation: $\frac{x^2}{9} + \frac{y^2}{4} = 1 \implies y = \frac{2}{3}\sqrt{9 - x^2}$ (taking the positive root for the first quadrant). Write the area: The rectangle has width $2x$ and height $2y$, so $A = (2x)(2y) = 4xy = 4x \cdot \frac{2}{3}\sqrt{9 - x^2} = \frac{8x}{3}\sqrt{9 - x^2}$. Why distractors fail: Option A ($\frac{4x}{3}\sqrt{9 - x^2}$) equals $2xy$, which only accounts for half the rectangle (e.g., using width $2x$ and height $y$ instead of $2y$). Option C ($\frac{2x}{3}\sqrt{9 - x^2}$) equals $xy$, the area of only one of the four congruent sub-rectangles in a single quadrant. Option D ($\frac{16x}{3}\sqrt{9 - x^2}$) overcounts by an extra factor of 2, as if the dimensions were $4x$ by $2y$.

Answer: $x = 30$ m, length $= 60$ m

Set up constraint and objective: Fencing: $2x + l = 120$, so $l = 120 - 2x$. Area: $A = x(120 - 2x) = 120x - 2x^2$. Find critical point: $A'(x) = 120 - 4x = 0 \implies x = 30$. Then $l = 120 - 60 = 60$. Verify maximum: $A''(x) = -4 < 0$, confirming a maximum at $x = 30$. Why distractors fail: Option A gives a square ($40 \times 40$) but uses fencing $2(40) + 40 = 120$, giving area $1600$ vs. the optimal $1800$. Option C gives area $1600$. Option D uses $180$ m of fencing, exceeding the constraint.

Answer: $x \approx 3.92$, because $x \approx 12.74$ exceeds the domain $0 < x < 10$

Determine the feasible domain: Since the sheet is $20 \times 30$ and we cut $x$ from each corner, we need $20 - 2x > 0$, so $x < 10$. Filter critical points: $x \approx 12.74 > 10$, so it is outside the domain and must be rejected. Only $x \approx 3.92$ is feasible. Why distractors fail: Option A selects the infeasible critical point. Option C does not check the domain. Option D is wrong because $x \approx 3.92$ is a valid solution.

Answer: About 44 cm

Set up the problem: Let $x$ be the wire for the circle and $100 - x$ for the square. Circle: circumference $= x$, so $r = \frac{x}{2\pi}$, area $= \pi r^2 = \frac{x^2}{4\pi}$. Square: perimeter $= 100 - x$, side $= \frac{100 - x}{4}$, area $= \frac{(100 - x)^2}{16}$. Total area and differentiation: $A(x) = \frac{x^2}{4\pi} + \frac{(100 - x)^2}{16}$. $A'(x) = \frac{x}{2\pi} - \frac{100 - x}{8} = 0$. Solving: $\frac{4x}{8\pi} = \frac{\pi(100 - x)}{8\pi}$, so $4x = \pi(100 - x)$, giving $x(4 + \pi) = 100\pi$, hence $x = \frac{100\pi}{4 + \pi} \approx 44.0$ cm. Why distractors fail: Options A and D are the boundary cases — these actually maximize total area (one can check that endpoints give larger total areas than the critical point). Option C splits equally but does not account for the different area efficiencies of circles and squares.

Answer: $f(1)$, $f(2)$, $f(7)$, and $f(10)$

Candidates Test for closed intervals: On a closed interval, the absolute maximum occurs either at a critical point or at an endpoint. All such candidates must be evaluated. Why the correct answer works: The student must compare $f(1)$, $f(2)$, $f(7)$, and $f(10)$. While $f''(7) < 0$ confirms a local max at $x = 7$, the absolute max could still be at an endpoint. Why distractors fail: Option A confuses local min ($f'' > 0$) with max. Option B identifies a local max but ignores endpoints. Option D ignores interior critical points.

Answer: The classmate used $x + l = P$ instead of $2x + l = P$ as the constraint

Correct constraint: With three sides of fencing: $2x + l = P$, so $l = P - 2x$ and $A = x(P - 2x)$. Then $A'(x) = P - 4x = 0 \implies x = P/4$. Identify the error: If the classmate wrote $x + l = P$, then $l = P - x$ and $A = x(P - x)$. Then $A'(x) = P - 2x = 0 \implies x = P/2$. This treats the two widths as one, counting only one side perpendicular to the wall. Why distractors fail: Option A describes the opposite error (including the wall). Option C blames calculus when the error is in the constraint setup. Option D misidentifies the objective function.