SOLUTION: A shopkeeper finds that the cost C of ordering and storing the latest laptop is C = 2 x + (300000 / x), where x is in the interval [1, 300]. The delivery of truck can bring at mos

Question 548215: A shopkeeper finds that the cost C of ordering and storing the latest laptop is
C = 2 x + (300000 / x), where x is in the interval [1, 300]. The delivery of truck can bring at most 300 units per
order.
(a) Find the order size that will minimize the cost
(b) Could the cost be decreased if the truck is replaced with a larger one that could bring 400 units?
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
A shopkeeper finds that the cost C of ordering and storing the latest laptop is
C = 2 x + (300000 / x), where x is in the interval [1, 300]. The delivery of truck can bring at most 300 units per
order.
(a) Find the order size that will minimize the cost
(b) Could the cost be decreased if the truck is replaced with a larger one that could bring 400 units?
=====================================
The minimum of the cost function is obtained by taking the derivative and setting=0
C(x) = 2x + 300000/x
dC(x)/dx = 0 = 2 - 300000/x^2
Divide through by 2, multiply through by x^2:
0 = x^2 - 150000
Taking the positive solution, we get x = sqrt(150000) = 387 [to the nearest unit]
(a) However, the truck can deliver at most 300 units. Therefore, the order size which will minimize the cost is 300 units.
(b) Yes. The cost is minimized when the number of units is 387
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