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For the given cost function C ( x ) = 32400 + 800 x + x 2 C(x)=32400+800x+x2 find:
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For the given cost function C ( x ) = 32400 + 800 x + x 2 C(x)=32400+800x+x2 find:
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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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
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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