Answer: Because multiplying the linear price function by $x$ produces a squared term

Trace where the quadratic term arises: Revenue $= p \times x = (50 - 2x) \times x = 50x - 2x^2$. The $x^2$ term appears because a linear expression in $x$ (the price) is multiplied by $x$ (the quantity). Why the correct answer works: Option B correctly identifies that the product of a linear function of $x$ and $x$ itself yields a degree-2 polynomial. Why distractors fail: Option A incorrectly attributes the quadratic nature to the cost function, which is not involved in computing revenue. Option C states the demand relationship is quadratic, but $p = 50 - 2x$ is linear. Option D makes an incorrect general claim; revenue is not always quadratic — it depends on the model.

Answer: Correct; the vertex form shows maximum profit of 100 dollars at $x = 10$ units

Verify by expansion: $-4(x-10)^2 + 100 = -4(x^2 - 20x + 100) + 100 = -4x^2 + 80x - 400 + 100 = -4x^2 + 80x - 300$. This matches the original expression. Interpret the vertex form: In $a(x-h)^2 + k$, the vertex is at $(h, k)$. Since $a = -4 < 0$, the parabola opens downward, so the vertex represents a maximum. Thus maximum profit is $k = 100$ at $x = h = 10$. Why distractors fail: Option A is wrong because the expansion does match. Option C incorrectly calls it a minimum; with $a < 0$, the vertex is a maximum. Option D uses $(x+10)$, which would shift the vertex to $x = -10$, a nonsensical quantity for units sold.

Answer: $R(x) = 12x$

Apply the revenue formula: Revenue equals price times quantity. Here, price is 12 dollars and quantity is $x$, so $R(x) = 12 \times x = 12x$. Why the correct answer works: Option B, $R(x) = 12x$, directly applies the formula Revenue = price × quantity with a constant price. Why distractors fail: Option A adds instead of multiplies. Option C incorrectly squares $x$, which has no basis in this model. Option D divides price by quantity, which is not how revenue is computed.

Answer: Model 1, with a maximum profit of 100 dollars

Find the maximum of Model 1: The vertex $x$-value is $x = -b/(2a) = -40/(2 \times -2) = 10$. Then $P_1(10) = -2(100) + 40(10) - 100 = -200 + 400 - 100 = 100$. Find the maximum of Model 2: The vertex $x$-value is $x = -54/(2 \times -3) = 9$. Then $P_2(9) = -3(81) + 54(9) - 150 = -243 + 486 - 150 = 93$. Compare and conclude: Model 1 predicts a maximum profit of 100 dollars, while Model 2 predicts 93 dollars. Model 1 is higher. Why distractors fail: Option B correctly computes Model 2's maximum as 93 but claims it is higher, which is wrong. Option C states Model 2's maximum is 100, which is a calculation error. Option D claims they are equal, but $100 \neq 93$.

Answer: Price decreases as quantity increases

Interpret the demand equation: In $p = 80 - 3x$, the coefficient of $x$ is $-3$, which is negative. This means that for each additional unit sold, the price drops by 3 dollars. Why the correct answer works: Option B correctly identifies that the negative slope causes price to decrease as quantity increases — a standard linear demand relationship. Why distractors fail: Option A describes a positive relationship, contradicting the negative coefficient. Option C describes a constant price (horizontal line), not a decreasing one. Option D describes an inverse proportionality ($p = k/x$), which is a hyperbola, not the linear equation given.

Answer: Profit = Revenue − Cost

Recall the profit definition: Profit is the amount remaining after subtracting all costs from total revenue. Why the correct answer works: Option C, Profit = Revenue − Cost, is the standard definition used in business and economics. Why distractors fail: Option A adds cost to revenue, which overstates earnings. Option B reverses the subtraction, giving a negative of the true profit. Option D multiplies revenue and cost, which has no meaningful financial interpretation.

Answer: $P(x) = -5x^2 + 90x - 200$

Find the revenue expression: $R(x) = p \times x = (100 - 5x)(x) = 100x - 5x^2$. Compute profit: $P(x) = R(x) - C(x) = (100x - 5x^2) - (200 + 10x) = -5x^2 + 100x - 10x - 200 = -5x^2 + 90x - 200$. Why the correct answer works: Option A correctly computes both revenue and then subtracts all cost terms. Why distractors fail: Option B adds the linear cost term instead of subtracting ($100x + 10x = 110x$). Option C fails to subtract the constant cost of 200 (gets $+200$ instead of $-200$). Option D reverses all signs, producing the negative of the correct profit.

Answer: The parabola opens downward, meaning revenue eventually decreases after a certain quantity

Interpret the leading coefficient: In a quadratic $ax^2 + bx + c$, if $a < 0$, the parabola opens downward. This means the function increases to a maximum and then decreases. Why the correct answer works: Option B correctly identifies that $a = -2 < 0$ produces a downward-opening parabola, so revenue rises to a peak and then falls — a realistic model for diminishing returns. Why distractors fail: Option A describes unbounded growth, which contradicts the negative leading coefficient. Option C confuses the coefficient of $x^2$ in the revenue expression with a per-unit cost. Option D confuses $a$ with the constant term; $R(0) = 0$, not $-2$.

Answer: $P(x) = -4x^2 + 120x + 3500$

Define price and quantity: Price $= 20 + x$. Quantity $= 200 - 4x$. Build the revenue expression: $R(x) = (20 + x)(200 - 4x) = 4000 - 80x + 200x - 4x^2 = -4x^2 + 120x + 4000$. Subtract fixed cost to get profit: $P(x) = R(x) - 500 = -4x^2 + 120x + 4000 - 500 = -4x^2 + 120x + 3500$. Why distractors fail: Option B forgets to subtract the fixed cost of 500 (leaving 4000). Option C uses $200x$ for the linear term, as if the quantity coefficient were not applied during expansion. Option D incorrectly makes the leading coefficient positive and mishandles the constant.

Answer: $a = -7,\; b = 70,\; c = -30$

Find revenue: $R(x) = (70 - 2x)(x) = 70x - 2x^2$. Compute profit: $P(x) = R(x) - C(x) = (70x - 2x^2) - (5x^2 + 30) = -2x^2 - 5x^2 + 70x - 30 = -7x^2 + 70x - 30$. Identify coefficients: So $a = -7$, $b = 70$, and $c = -30$. Why distractors fail: Option B uses $a = -3$, which would result from subtracting incorrectly ($-2 - 5 \neq -3$). Option C gets $c = 30$ instead of $-30$ (sign error on the constant). Option D reverses all signs.

Answer: No; the constant term should be $-12$, not $+12$

Compute the correct profit: $P(x) = R(x) - C(x) = (-x^2 + 20x) - (4x + 12) = -x^2 + 20x - 4x - 12 = -x^2 + 16x - 12$. Identify the error: The consultant wrote $+12$ instead of $-12$. This is a sign error when subtracting the constant cost term. Why distractors fail: Option A is wrong because the consultant made a sign error on the constant. Option C is wrong because $20x - 4x = 16x$ is correctly computed. Option D is wrong because the $-x^2$ term is carried directly from $R(x)$ with no change needed.

Answer: Both expressions are equivalent; rearranging terms does not change the value

Recognize equivalent expressions: $30x - x^2$ and $-x^2 + 30x$ contain the same terms; addition is commutative, so their order does not affect the expression's value. Why the correct answer works: Option C correctly states that both forms are algebraically identical. Why distractors fail: Option A incorrectly suggests term order matters for correctness. Option B implies only one ordering is valid, which is false — standard form is a convention, not a requirement for correctness. Option D reverses both signs, producing an entirely different expression.

Answer: $R(x) = 40x - x^2$

Apply Revenue = price × quantity: $R(x) = p \times x = (40 - x) \times x = 40x - x^2$. Why the correct answer works: Option A, $R(x) = 40x - x^2$, is the correct expansion of $(40 - x)(x)$. Why distractors fail: Option B is missing the proper linear term; it appears that only the $-x$ was multiplied by $x$ while the constant 40 was not. Option C uses a positive $x^2$ term, which would come from $(40 + x)(x)$, not $(40 - x)(x)$. Option D incorrectly places the exponent, yielding $40x^2 - x$ which does not follow from the given price function.

Answer: $P(x) = -3x^2 + 50x - 50$

Set up the profit expression: $P(x) = R(x) - C(x) = (60x - 3x^2) - (10x + 50)$. Distribute the negative sign and simplify: $P(x) = 60x - 3x^2 - 10x - 50 = -3x^2 + 50x - 50$. Why the correct answer works: Option B correctly subtracts every term in the cost function, giving $-3x^2 + 50x - 50$. Why distractors fail: Option A adds the linear cost term instead of subtracting it ($60x + 10x = 70x$). Option C gets the linear term right but fails to subtract the constant 50 (sign error on constant). Option D flips the sign on the $x^2$ term, making it positive.