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Science & Mathematics by Anonymous 2018-07-29 06:45:51
Social Science
An open top box is made from a square base with an area of 100 square feet by cutting equal?
4 answers
squares off each corner, and then folding the sides up. What dimensions of the box would maximize its volume? is it 20/3 ft?
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Anonymous
So we know that the base is a square and its area is 100 square feet. That means that the base is 10ft by 10ft. We're going to cut a length of x from each corner, which means that we will cut 2x from each side. Folding the sides up, we now have this formula for the volume: V = (10 - 2x) * (10 - 2x) * x V = (10 - 2x)^2 * x V = 2^2 * (5 - x)^2 * x V = 4 * (5 - x)^2 * x V = 4 * (25 - 10x + x^2) * x V = 4 * (25x - 10x^2 + x^3) Now derive and let dV/dx = 0 dV/dx = 4 * (25 - 20x + 3x^2) dV/dx = 0 0 = 4 * (25 - 20x + 3x^2) 0 = 25 - 20x + 3x^2 0 = 3x^2 - 20x + 25 x = (20 +/- sqrt(400 - 4 * 3 * 25)) / 6 x = (20 +/- sqrt(100 * (4 - 3))) / 6 x = (20 +/- 10 * sqrt(1)) / 6 x = 10 * (2 +/- 1) / 6 x = 10 * (3/6) , 10 * (1/6) x = 5 , 5/3 If we cut 5 ft from each corner, we'll have a volume of 0. If we cut 5/3 ft from each corner, we'll have a max volume V = (10 - 2x)^2 * x V = (10 - 2 * (5/3))^2 * (5/3) V = (5/3) * (10 - 10/3)^2 V = (5/3) * (20/3)^2 V = (5/3) * (400/9) V = 2000/27 The maximum volume is 2000/27 cubic feet The portion cut from the corners is 5/3 feet, or 1 ft 8 inches
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Anonymous
20/3 is right. You can check if true: If you cut off squares of 10/6 feet on a side, dimensions are 20/3 x 20/3 x 10/6. That is a volume of 4000/54, or a about 74.07 cubic feet. If you cut squares of 10/5.9 feet on a side, the dimensions are 6.610 x 6.610 x 1.695 and volume is 74.06 or less than your number If you cut squares of 10/6.1 feet on a side, the dimensions are 6.721 x 6.721 x 1.693 and volume is 74.056 or less than your number
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Anonymous
square base is 10 ft by 10 ft cutouts are x ft on each side, x < 5 ft dimensions of box: length = 10-2x ft, width = 10-2x ft, height = x ft volume V = x(10-2x)² = x(4x²-40x+100) = 4x³ - 40x² + 100x max volume when first derivative is 0 dV/dx = 12x² - 80x + 100 = 0 Quadratic formula x = [80 ± √(80² – 4·12·100)] / [2·12] = [80 ± √1600] / 24 = [80 ± 40] /24 = 5/3, 5 x=5 is an extraneous solution x = 5/3 ft dimensions of box: length = 20/3 ft, width = 20/3 ft, height = x ft
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Anonymous
V = x(10-2x)^2 V = 100x-40x^2+4x^3 = 0 dV/dx = 100-80x+12x^2 = 0 3x^2-20x+25 = 0 (3x-5)(x-5) = 0 x = 5/3 ft Height = 5/3 ft Base = 20/3 ft by 20/3 ft