The problem provides the cost function $C(x) = -2350 + 80x - 0.04x^2$ and the price function $P(x) = 2x - 35$ for a company that sells kids' toys. The problem asks us to: (a) Find the marginal profit function. (b) Determine the level of production that maximizes the profit, rounded to the nearest whole number. (c) Calculate the maximum annual profit, rounded to the nearest whole number.

1. Problem Description

The problem provides the cost function $C(x) = -2350 + 80x - 0.04x^2$ and the price function $P(x) = 2x - 35$ for a company that sells kids' toys. The problem asks us to:

(a) Find the marginal profit function.

(b) Determine the level of production that maximizes the profit, rounded to the nearest whole number.

(c) Calculate the maximum annual profit, rounded to the nearest whole number.

2. Solution Steps

(a) To find the marginal profit function, we first need to find the revenue function $R(x)$ and the profit function $Pr(x)$.

The revenue function is given by:

$R(x) = x \cdot P(x)$

$R(x) = x(2x - 35) = 2x^2 - 35x$

The profit function is given by:

$Pr(x) = R(x) - C(x)$

$Pr(x) = (2x^2 - 35x) - (-2350 + 80x - 0.04x^2)$

$Pr(x) = 2x^2 - 35x + 2350 - 80x + 0.04x^2$

$Pr(x) = 2.04x^2 - 115x + 2350$

The marginal profit function is the derivative of the profit function:

$Pr'(x) = \frac{d}{dx} (2.04x^2 - 115x + 2350)$

$Pr'(x) = 4.08x - 115$

(b) To maximize profit, we need to find the critical points by setting the marginal profit function equal to zero and solving for $x$:

$Pr'(x) = 4.08x - 115 = 0$

$4.08x = 115$

$x = \frac{115}{4.08} \approx 28.186$

Since we need to round the answer to the nearest whole number, $x \approx 28$.

To verify that this is a maximum, we can take the second derivative of the profit function:

$Pr''(x) = \frac{d^2}{dx^2} Pr(x) = \frac{d}{dx}(4.08x - 115) = 4.08$

Since $Pr''(x) = 4.08 > 0$, the profit function is concave up, and thus $x = 28$ is a minimum. There seems to be a mistake in the equation for $C(x)$ because $Pr''(x)$ must be negative to indicate a maximum. Based on the original image, it seems the problem is written as such. I suspect the problem intended for $C(x) = -2350 + 80x + 0.04x^2$. If we make that change, here are the new calculations:

$Pr(x) = (2x^2 - 35x) - (-2350 + 80x + 0.04x^2)$

$Pr(x) = 2x^2 - 35x + 2350 - 80x - 0.04x^2$

$Pr(x) = 1.96x^2 - 115x + 2350$

$Pr'(x) = 3.92x - 115$

$3.92x - 115 = 0$

$x = \frac{115}{3.92} \approx 29.3367$

Since we need to round the answer to the nearest whole number, $x \approx 29$.

$Pr''(x) = 3.92 > 0$ so this would still be a minimum.

I suspect there is an error somewhere.

If the question had instead asked to minimize costs, then the answer would be x = 28 or

2

9. Assuming that the question actually intended the following:

C(x) = 2350 + 80x + 0.04x^2

and

P(x) = 120 - 2x

Then

R(x) = (120 - 2x)x = 120x - 2x^2

Pr(x) = R(x) - C(x)

Pr(x) = 120x - 2x^2 - (2350 + 80x + 0.04x^2)

Pr(x) = -2.04x^2 + 40x - 2350

Pr'(x) = -4.08x + 40 = 0

x = 40 / 4.08 = 9.80

Round to nearest whole number gives x = 10

Pr''(x) = -4.08 < 0 so this is indeed a maximum.

Then part (c) would give

Pr(10) = -2.04*(10)^2 + 40*(10) - 2350 = -204 + 400 - 2350 = -2104

We will proceed assuming C(x) = -2350 + 80x - 0.04x^2 and P(x) = 2x - 35

and therefore Pr(x) = 2.04x^2 - 115x + 2350

(c) Since we found in (b) that we actually have a minimum instead of a maximum, we need to look at the endpoints of a feasible domain. Assume that we want $x > 0$ and $P(x) = 2x - 35 > 0$, so $x > 17.5$, or $x \ge 18$.

Then consider large values for $x$, and profit is growing.

The question makes no sense, since profit is not maximized, but minimized.

3. Final Answer

(a) The marginal profit function is $Pr'(x) = 4.08x - 115$.

(b) The level of production that minimizes the profit is approximately

2

8. (c) Assuming that the level of production is 28, the annual profit is:

$Pr(28) = 2.04(28)^2 - 115(28) + 2350 = 2.04(784) - 3220 + 2350 = 1599.36 - 3220 + 2350 = 729.36 \approx 729$

If the level of production is 29, the annual profit is:

$Pr(29) = 2.04(29)^2 - 115(29) + 2350 = 2.04(841) - 3335 + 2350 = 1715.64 - 3335 + 2350 = 730.64 \approx 731$

Assuming there are no bounds on the number of toys. Since the profit is minimized near 28-29 toys, the maximum profits will be generated at a much higher production level.