An open box problem

Question

an open box is to be made from a 15 ft by 40 ft rectangular piece of sheet metal by cutting out squares of equal size from the four corners and bending up the sides. find the maximum volume that the box can have.

Answer 1

Draw a diagram to confirm that the dimensions of the 3-d box will be V(x) = x(15 - 2x)(40 - 2x) , where x is the side length of the squares to be cut from the four corners. Because 15 is the smaller of the two given dimensions, the valid domain for the volume function is 0 < x < 7.5. We find the maximum volume by taking the derivative (using power rule), setting it = 0, and solving. That gives the ideal x-value, which we finally plug into the volume equation to arrive at the max volume:

V(x) = 4x³ - 110x² + 600x ; 0 < x < 7.5

V^'(x) = 12x² - 220x + 600 = 0

3x² - 55x + 150 = 0 Use quadratic formula to proceed ...