Maximize Volume of a Box
Optimization Problem
How to maximize the volume of a box using the first derivative of the volume. A volume optimization problem with solution.
Problem
A sheet of metal 12 inches by 10 inches is to be used to make a open box. Squares of equal sides x are cut out of each corner then the sides are folded to make the box. Find the value of x that makes the volume maximum.
Solution to Problem 1:
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We first use the formula of the volume of a rectangular box.
V = L × W × H
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The box to be made has the following dimensions:
L = 12 - 2 x
W = 10 - 2 x
H = x
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We now write the volume of the box to be made as follows:
V(x) = x (12 - 2 x) (10 - 2 x) = 4 x (6 - x) (5 - x)
= 4x (x 2 -11 x + 30)
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We now determine the domain of function V(x). All dimensions of the box must be positive or zero, hence the conditions
x ≥ 0 and 6 - x ≥ 0 and 5 - x ≥ 0
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Solve the above system of inequalities to find the domain of function V(x)
0 ≤ x ≤ 5
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Let us now find the first derivative of V(x) using its last expression.
dV / dx = 4 [ (x 2 -11 x + 3) + x (2x - 11) ]
= 3 x 2 -22 x + 30
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Let us now find all values of x that make dV / dx = 0 by solving the quadratic equation
3 x 2 -22 x + 30 = 0
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Two values make dV / dx = 0:
x = 5.52 and x = 1.81, rounded to one decimal place.
x = 5.52 is outside the domain and is therefore rejected.
Let us now examine the values of V(x) at x = 1.81 and the endpoints of the domain.
V(0) = 0 , v(5) = 0 and V(1.81) = 96.77 (rounded to two decimal places)
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So V(x) is maximum for x ≈ 1.81 inches. The graph of function V(x) is shown below and we can clearly see that there is a maximum very close to 1.8.
More references on
calculus problems
