The marginal cost in dollars of producing x units is given by the next equation:
[tex]C=2.6\sqrt[]{x}+600[/tex]
a)
To find the marginal cost (in dollars per unit) when x= 9.
Then, we need to replace x=9 on the derivation of the cost equation:
So:
[tex]\frac{d}{dx}C=\frac{1.3}{\sqrt[]{x}}[/tex]
Where:
[tex]\frac{d}{dx}2.6\sqrt[]{x}=2.6\frac{d}{dx}\sqrt[]{x}=2.6\frac{d}{dx}^{}x^{\frac{1}{2}}=2.6\cdot\frac{1}{2}x^{\frac{1}{2}-1}=1.3\cdot x^{-\frac{1}{2}}=\frac{1.3}{\sqrt[]{3}}[/tex]
and, the derivate of a constant is equal to zero.
[tex]\frac{d}{dx}600=0[/tex]
Replacing x= 9
[tex]\frac{d}{dx}C=\frac{1.3}{\sqrt[]{9}}[/tex]
Hence, the marginal cost is equal to:
[tex]\frac{d}{dx}C=0.43[/tex]
b) Now, when the production increases 9 to 10. It's the same as the cost of producing one more machine beyond 9.
Then, it would be x=10 on the cost equation:
[tex]C=2.6\sqrt[]{x}+600[/tex][tex]C=2.6\sqrt[]{10}+600[/tex][tex]C=608.22[/tex]
and x= 9
[tex]C=2.6\sqrt[]{9}+600[/tex][tex]C=2.6(3)+600[/tex][tex]C=607.8[/tex]
Then, we calculate C(10) - C(9) =
[tex]608.22-607.8[/tex][tex]=0.43[/tex]
C)
Both results are equal.
Hence, the marginal cost when x=9 is equal to the additional cost when the production increases from 9 to 10.
