Answer:
The dimensions of the square, that will result in a box of maximum volume is a square with side length of approximately 1.585 inches
Step-by-step explanation:
The given dimension of the steel sheet are;
The width = 8.5 inches
The length = 11 inches
Let 'x' represent the side length of the squares cut out, we have;
The length of the box formed, l = 11 - 2·x
The width of the box formed, w = 8.5 - 2·x
The height of the box, h = x
The volume of the box, V = (11 - 2·x) × (8.5 - 2·x) × (x) = 4·x³ - 39·x² + 93.5·x
The maximum or minimum volume is found from dV/dx = 0 at max. or min.
dV/dx = d(4·x³ - 39·x² + 93.5·x)/dx = 12·x² - 78·x + 93.5 = 0
Which gives;
x = (78 ± √((-78)² - 4×12×93.5))/(2 ×12)
x = 4.915 or 1.585
When x = 4.915, V = 4 × 4.915³ - 39 × 4.915² + 93.5 × 4.915 = -7.65
When x = 1.585, V = 4 × 1.585³ - 39 × 1.585² + 93.5 × 1.585 = 66.15
Therefore, the local maximum occurs at x ≈ 1.585
Therefore, the dimensions of the square, 'x' that will result in a box of maximum volume is x ≈ 1.585 inches.
