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 $\bar { C } ( x ) = 5 x + \frac { 4500 } { x } , \quad x 0$ , where x is the number of machines used. What is the minimum average cost?
A)$0
B)$30
C)$300
D)$150
E)$4505

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 $\bar { C } ( x ) = 5 x + \frac { 4500 } { x } , \quad x > 0$ , where x is the number of machines used. What is the minimum average cost?
A)$0
B)$30
C)$300
D)$150
E)$4505

Quizplus · Accepted Answer

To find the minimum average cost, we differentiate the given average cost function with respect to $x$ and set the derivative equal to zero. The average cost function is $\bar { C } ( x ) = 5 x + \frac { 4500 } { x }$. Differentiating with respect to $x$, we get $\frac { d \bar { C } } { d x } = 5 - \frac { 4500 } { x^2 }$. Setting this equal to zero to find the critical point gives $5 = \frac { 4500 } { x^2 }$, which simplifies to $x^2 = 900$, so $x = 30$. Substituting $x = 30$ back into the original average cost function gives the minimum average cost: $\bar { C } ( 30 ) = 5(30) + \frac { 4500 } { 30 } = 150 + 150 = 300$.

Average Costs Cˉ(x)=5x+4500x,x>0\bar { C } ( x ) = 5 x + \frac { 4500 } { x } , \quad x > 0Cˉ(x)=5x+x4500​,x>0

Average Costs $\bar { C } ( x ) = 5 x + \frac { 4500 } { x } , \quad x > 0$