Suppose the dollar cost of producing x video cameras is C(x)=500x−0

In order to complete this problem, we'll need to use the given cost function to produce two additional functions: the marginal cost and average cost functions. The marginal cost function can be constructed by finding the derivative of the given function, and the average cost function can be constructed by dividing the given function by x.

C′(x)=500−0.018x+1300000000x2¯C(x)=500x−0.009x2+(10−8)(x3)x=500−0.09x+1100000000x2C′(x)=500−0.018x+1300000000x2C¯(x)=500x−0.009x2+(10−8)(x3)x=500−0.09x+1100000000x2

First, let's find the marginal cost when 6500 video cameras are produced. This can be found by substituting in 6500 to the derivative of the cost function.

C′(6500)=500−0.018(6500)+1300000000(6500)2=384.267C′(6500)=500−0.018(6500)+1300000000(6500)2=384.267

This means that the cost to produce the next camera, the 6501st camera, can be approximated to be $384.27. The actual cost can be found by finding the difference of producing 6501 and 6500 cameras.

C(6500)=500(6500)−0.009(6500)2+(10−8)((6500)3)=2872496.25C(6501)=500(6501)−0.009(6501)2+(10−8)((6501)3)=2872880.509C(6501)−C(6500)=2872880.509−2872496.25=384.259C(6500)=500(6500)−0.009(6500)2+(10−8)((6500)3)=2872496.25C(6501)=500(6501)−0.009(6501)2+(10−8)((6501)3)=2872880.509C(6501)−C(6500)=2872880.509−2872496.25=384.259

Thus, the actual cost of producing the 6501st camera is $384.26. Next, let's find the average cost per camera when 6500 are produced.

¯C(6500)=500−0.09(6500)+1100000000(6500)2=441.9225C¯(6500)=500−0.09(6500)+1100000000(6500)2=441.9225

Therefore, when 6500 cameras are produced, each costs approximately $441.92 to make.