If 1200 cm2 of material is available to make a box with a sq

Solution

We have 1200 cm^2 worth of material. This represents the surface area of a box with a square base and an open top. the dimensions of the box. Square base implies the length and width are the same. Let x = length/width. The box has a height as well. Let h = height the surface area ; S. S = (area of base) + (area of 4 walls) The area of the base is x^2 The area of one of the walls is length times height, or xh. Since there are 4 of them, it would be 4 times xh, or 4xh. S = x^2 + 4xh And we know that S = 1200cm^2, so x^2 + 4xh = 1200 Let\'s solve for h. 4xh = 1200 - x^2 h = (1200 - x^2) / (4x) we require the volume formula. V = (length) x (width) x (height) And we know all of these. V = (x)(x)(h) V = (x^2) h putting h = (1200 - x^2) / (4x) in the formula V = (x^2) ( 1200 - x^2)/(4x) We get a cancellation, V = x(1200 - x^2)/4 V = (1/4)x (1200 - x^2) This will be our volume function, V(x). V(x) = (1/4)(x)(1200 - x^2) To maximize V(x), we must first take the derivative and then make it 0. Using the product rule (and ignoring the constant 1/4), we have V\'(x) = (1/4) [ (1200 - x^2) + (x)(-2x) ] Simplify, V\'(x) = (1/4) [ 1200 - x^2 - 2x^2 ] V\'(x) = (1/4) [ 1200 - 3x^2 ] V\'(x) = (1/4) [ 3(400 - x^2) ] V\'(x) = (3/4) [ 400 - x^2 ] To maximize, make V\'(x) = 0, and solve for x. 0 = (3/4) [ 400 - x^2 ] 0 = 400 - x^2 x^2 = 400 x = +/- 20 Therefore, x = { 20, -20 } However, since x represents a dimension, it can never be negative, and we must discard the negative solution. That means x = 20. This tells us that the maximum volume occurs when x = 20. However, the question is asking WHAT the largest volume of the box is. Solving this is as simple as plugging x = 20 into our volume function, V(x). V(x) = (1/4)(x)(1200 - x^2) Therefore, V(20) = (1/4) (20) (1200 - 20^2) V(20) = (1/4) (20) (1200 - 400) V(20) = (1/4) (20) (800) V(20) = (20/4)(800) V(20) = 5(800) V = 4000cm^3