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Express Lane Plus
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3 weeks ago
For the given cost function

C ( x ) = 32400 + 800 x + x 2

a) The production level that will minimize the average cost
b) The minimal average cost
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3 weeks ago
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Ok, so they're asking for the maximum.

Take the derivative.

C'( x ) = 800 + 2x

Now set c'(x) = 0 to find the critical point(s)

0 = 800 + 2x
-800 = 2x
-800/2 = -400 = x

When x = -400, that's when you're get the maximum production.


Whoops, they want the minimum of the average cost Grinning Face with Smiling Eyes

The average cost is total cost divided by number of products or C(x)/x.

\(A\left(x\right)=\frac{C\left(x\right)}{x}=\frac{\left(32400+800x+x^2\right)}{x}\)

\(A\left(x\right)=\frac{32400}{x}+800+x\)

Find the derivative of this, then set to 0 to find the critical points.

\(A'\left(x\right)=\frac{-32400}{x^2}+1\)

\(0=\frac{-32400}{x^2}+1\)

\(-1=\frac{-32400}{x^2}\)

\(x^2=32400\)

\(x=\pm 180\) (critical points) -- production level that will minimize the cost is 180.

Substitute these in A(x)

\(A\left(180\right)=\frac{32400}{180}+800+180=1160\)

Minimal cost is 1160
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