a regular sheet of letter (printing) paper measrues 8.5 inches by 11 inches. you want to cut out squares from each corner that it may be folded up to produce an open - top box. What are the dimensions of the square cut outs that will result in a box of maximum volume

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.

area = length x width

length = 2 x width

32 = 2 x width x width

32 = 2 x width^2

36 = width ^2  

width = 6

length = 12

area is 32

perimeter = 2 x 12 + 2 x 6 = 24 + 12 = 36

step-by-step explanation:

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step-by-step explanation: