A shipping company must design a closed rectangular shipping crate with a square base. The volume is 1920ft31920ft3. The material for the top and sides costs $2$2 per square foot and the material for the bottom costs $13$13 per square foot. Find the dimensions of the crate that will minimize the total cost of material.
Question
A shipping company must design a closed rectangular shipping crate with a square base. The volume is 1920ft31920ft3. The material for the top and sides costs 2 per square foot and the material for the bottom costs 13 per square foot. Find the dimensions of the crate that will minimize the total cost of material.
Solution
To solve this problem, we need to use calculus and the method of optimization. Here are the steps:
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Let's denote the side length of the square base as x (in feet) and the height of the crate as h (in feet). The volume V of the crate is given by V = x^2 * h. We know that V = 1920 ft^3, so we can express h in terms of x: h = 1920 / x^2.
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The total cost C of the material is the sum of the cost of the top, sides, and bottom. The top and sides are made of the same material, so their cost is 2 * (area of top + area of sides) = 2 * (x^2 + 4xh). The bottom is made of a different material, so its cost is 13 * (area of bottom) = 13 * x^2.
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Substituting h = 1920 / x^2 into the cost equation, we get C = 2 * (x^2 + 4x1920/x^2) + 13 * x^2 = 2x^2 + 15360/x + 13x^2.
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To minimize C, we take the derivative of C with respect to x and set it equal to zero: C' = 4x - 15360/x^2 + 26x = 0. Solving this equation for x, we get x = 30 ft.
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Substituting x = 30 ft into the equation h = 1920 / x^2, we get h = 1920 / 900 = 2.13 ft.
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So, the dimensions that will minimize the total cost of material are a base of 30 ft by 30 ft and a height of 2.13 ft.
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