# Thread: Calc - Word problem

1. ## Calc - Word problem

A special CD is manufactured and its manufacturing costs is given by the equation C(x) = 0.08x^2 + 0.2x + 3200. If the price function is given by the equation p(x) = 33 - 0.02x, determine the price that should be charged

a) to maximize profits

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

$R(x) = \frac {33 - 0.02x} {x}$

$R(x) = \frac {33}{x} - 0.02$

$P(x) = \frac {33}{x} - 0.02 - 0.08x^2 - 0.2x - 3200$

$P(x) = \frac {33}{x} -3200.02 - 0.2x - 0.08x^2$

$P'(x) = \frac {33}{x^2} -3200.02 - 0.2x - 0.08x^2$

$0 = \frac {33}{x^2} -3200.02 - 0.2x - 0.08x^2$

$3200.02 = \frac {33}{x^2} - 0.2x - 0.08x^2$

$3200.02 = \frac {33 - 0.2x^{3} - 0.008^{4}}{x^2}$

$3200.02x^2 = 33 - 0.2x^{3} - 0.008^{4}$

$3200.02x^2 = 33 - 0.2x^{3} - 0.008^{4}$

Now I don't know what to do. . .? I can't seem to find the zeros of the first derivative. . .

2. What does x represent?

3. Let x represent the number of CDs to maximize profit

4. Originally Posted by Macleef
A special CD is manufactured and its manufacturing costs is given by the equation C(x) = 0.08x^2 + 0.2x + 3200. If the price function is given by the equation p(x) = 33 - 0.02x, determine the price that should be charged

a) to maximize profits

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

$R(x) = \frac {33 - 0.02x} {x}$

$R(x) = \frac {33}{x} - 0.02$

$P(x) = \frac {33}{x} - 0.02 - 0.08x^2 - 0.2x - 3200$

$P(x) = \frac {33}{x} -3200.02 - 0.2x - 0.08x^2$

$P'(x) = \frac {33}{x^2} -3200.02 - 0.2x - 0.08x^2$

$0 = \frac {33}{x^2} -3200.02 - 0.2x - 0.08x^2$

$3200.02 = \frac {33}{x^2} - 0.2x - 0.08x^2$

$3200.02 = \frac {33 - 0.2x^{3} - 0.008^{4}}{x^2}$

$3200.02x^2 = 33 - 0.2x^{3} - 0.008^{4}$

$3200.02x^2 = 33 - 0.2x^{3} - 0.008^{4}$

Now I don't know what to do. . .? I can't seem to find the zeros of the first derivative. . .

$P(x)= \frac {33}{x} -3200.02 - 0.2x - 0.08x^2$

$P'(x)=-\frac{33}{x^2}-0.2-0.16x$

Setting equal to zero, we get:

$0=33+0.2x^2+0.16x^3$ (I multiplied through by $x^2$)

I would recommend using your calculator to find the values of x

I get $x\approx -6.63$.

This doesn't seem right. Did you used the correct formula for revenue? Shouldn't it be $R(x)=x\cdot P(x)$??

When $R(x)=x\cdot P(x)$, I get $x\approx 59$ CDs.

This is more reasonable.

5. Yes you are correct, revenue is x(px) ... I was thinking of the average cost formual -_-

and when I took the derivative of the new formula, I get 164 CDs

6. Originally Posted by Macleef
Yes you are correct, revenue is x(px) ... I was thinking of the average cost formual -_-

and when I took the derivative of the new formula, I get 164 CDs
What did you get for the new profit function? I get $P(x)=-0.28x^2+32.8x-3200$. Thus, $P'(x)=-0.56x+32.8$.

Setting equal to zero, I get $x=\frac{32.8}{.56}\approx 58.57 \implies 59$ CDs.