A manufacturer has determined that the total cost [tex]$C$[/tex] of operating a factory is
[tex]$C=0.5 x^2+15 x+5000$[/tex]
where [tex]$x$[/tex] is the number of units produced. At what level of production will the average cost per unit be minimized? (The average cost per unit is [tex]$\frac{C}{x}$[/tex]). [tex]$x=$[/tex] units
Answer (1)
Define the average cost function: A ( x ) = 0.5 x + 15 + x 5000 .
Find the first derivative: A'(x) = 0.5 - \\\\[\frac{5000}{x^2}\\\] .
Set the first derivative to zero and solve for x : 0.5 - \\\\[\frac{5000}{x^2}\\\] = 0 , which gives x = 100 .
Verify that this is a minimum by checking the second derivative: 0"> A''(x) = \\\\[\frac{10000}{x^3}\\\] > 0 at x = 100 .
The level of production that minimizes the average cost per unit is 100 units.
Explanation
Problem Analysis We are given the total cost function C = 0.5 x 2 + 15 x + 5000 , where x is the number of units produced. We want to find the level of production x that minimizes the average cost per unit, which is given by x C .
Finding the Average Cost Function The average cost per unit, A ( x ) , is given by A ( x ) = x C = x 0.5 x 2 + 15 x + 5000 = 0.5 x + 15 + x 5000 .
Finding the Derivative To minimize the average cost, we need to find the critical points of A ( x ) . We do this by taking the derivative of A ( x ) with respect to x and setting it equal to zero: A ′ ( x ) = d x d ( 0.5 x + 15 + x 5000 ) = 0.5 − x 2 5000 .
Solving for Critical Points Now, we set A ′ ( x ) = 0 and solve for x :
0.5 − x 2 5000 = 0 ⇒ 0.5 = x 2 5000 ⇒ x 2 = 0.5 5000 = 10000 ⇒ x = ± 10000 = ± 100. Since x represents the number of units produced, it must be positive. Therefore, x = 100 .
Finding the Second Derivative To ensure that x = 100 corresponds to a minimum, we compute the second derivative of A ( x ) :
A ′′ ( x ) = d x 2 d 2 ( 0.5 x + 15 + x 5000 ) = d x d ( 0.5 − x 2 5000 ) = x 3 10000 .
Verifying Minimum Now, we evaluate A ′′ ( x ) at x = 100 :
0."> A ′′ ( 100 ) = 10 0 3 10000 = 1000000 10000 = 0.01 > 0. Since 0"> A ′′ ( 100 ) > 0 , the average cost is minimized at x = 100 .
Final Answer Therefore, the level of production that minimizes the average cost per unit is x = 100 units.
Examples
In business, determining the production level that minimizes average cost is crucial for maximizing profit margins. For example, a bakery can use this calculation to find the number of cakes they need to produce each day to minimize the average cost per cake, taking into account fixed costs like rent and equipment, and variable costs like ingredients and labor. By optimizing production, the bakery can offer competitive prices while maintaining profitability. This principle applies across various industries, from manufacturing to service-based businesses, highlighting the importance of cost optimization in achieving business success.
