Problem 74 - College Algebra and Calculus: An Applied Approach

Question

The cost \(C\) for ordering and storing \(x\) units is \(C=2 x+300,000 / x .\) What order size will produce a minimum cost?

Step-by-Step Solution

Answer

The order size that will result in a minimum cost is \(\sqrt{150,000} units\).

Step 1: Find the derivative of the cost function

First, find the derivative of the cost function. The derivative of \(C(x) = 2x + \frac{300,000}{x}\) is \(C'(x) = 2 - \frac{300,000}{x^2}\), which results from applying the Power Rule to the first term and the Quotient rule to the second term

Step 2: Determine the critical point

Next, find the value of \(x\) that will yield a zero derivative by setting \(C'(x)\) equal to zero. This gives \(2 - \frac{300,000}{x^2} = 0\). Solving for \(x\), we obtain \(x = \sqrt{\frac{300,000}{2}} = \sqrt{150,000}\)

Step 3: Ensure that the critical point yields a minimum value

Now, we need to confirm if \(x = \sqrt{150,000}\) indeed results to a minimum cost. This can be done by either calculating the second derivative or reasoning based on the nature of the cost function \(C(x) = 2x + \frac{300,000}{x}\) which is always positive. Since ordering less than or more than \(\sqrt{150,000}\) units will result in added costs due to under-utilising the fixed cost or due to increased marginal ordering cost, it can be reasoned that \(x = \sqrt{150,000}\) is indeed a minimum point.