Calc - Word problem

**Macleef** · June 2nd 2008, 02:42 PM

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. . .

**sean.1986** · June 2nd 2008, 03:11 PM

What does x represent?

**Macleef** · June 2nd 2008, 11:35 PM

Let x represent the number of CDs to maximize profit

**Chris L T521** · June 3rd 2008, 12:05 AM

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. . .

You differentiated wrong...I'll help you through this.

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

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

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.

**Macleef** · June 3rd 2008, 12:17 AM

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

**Chris L T521** · June 3rd 2008, 12:23 AM

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.

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