Average costs. Suppose the average costs of a mining operation depend on the number of machines used, and average costs, in dollars, are given by $\bar { C } ( x ) = 7 x + \frac { 1372 } { x } , \quad x 0$ , where x is the number of machines used. How many machines give minimum average costs?
A)Using 14 machines gives the minimum average costs.
B)Using zero machines gives the minimum average costs.
C)Using 24 machines gives the minimum average costs.
D)Using 28 machines gives the minimum average costs.
E)Using 33 machines gives the minimum average costs.

Question

Average costs. Suppose the average costs of a mining operation depend on the number of machines used, and average costs, in dollars, are given by $\bar { C } ( x ) = 7 x + \frac { 1372 } { x } , \quad x > 0$ , where x is the number of machines used. How many machines give minimum average costs?
A)Using 14 machines gives the minimum average costs.
B)Using zero machines gives the minimum average costs.
C)Using 24 machines gives the minimum average costs.
D)Using 28 machines gives the minimum average costs.
E)Using 33 machines gives the minimum average costs.

Quizplus · Accepted Answer

To find the minimum average cost, we differentiate the given function $\bar { C } ( x ) = 7 x + \frac { 1372 } { x }$ with respect to $x$ and set the derivative equal to zero. The derivative is $\bar { C }' ( x ) = 7 - \frac { 1372 } { x^2 }$. Setting this equal to zero gives $7 = \frac { 1372 } { x^2 }$, solving for $x$ gives $x = \sqrt{ \frac { 1372 } { 7 } } = \sqrt{ 196 } = 14$. Therefore, using 14 machines gives the minimum average costs.

Average Costs Cˉ(x)=7x+1372x,x>0\bar { C } ( x ) = 7 x + \frac { 1372 } { x } , \quad x > 0Cˉ(x)=7x+x1372​,x>0

Average Costs $\bar { C } ( x ) = 7 x + \frac { 1372 } { x } , \quad x > 0$