What will the dimensions of the resulting cardboard box be if the company wants to maximize the volume and they start with a flat piece of square cardboard 20 feet per side, and then cut smaller squares out of each corner and fold up the sides to create the box?

Question

Luke Alvarez · Accepted Answer

Suppose that the squares removed from each corner are #x# feet by #x# feet each. When these are folded up they give a box with a height of #x# feet
and a base of #20 - 2x# feet by #20-2x# feet
for a volume
#V = x(20-2x)^2= 400x -80x^2 + 4x^3# To find the critical point(s) take the derivative of #V#, set it to zero, and solve for #x#. #(dV)/(dx) = 400-160x+12x^2# #=4(3x-10)(x-10) = 0# Since #x=10# gives a Volume of #0# #rarr# the critical point for the Volume that is it's maximum occurs when #x=10/3#. The resulting box will be
#3 1/3 xx 13 1/3 xx 13 1/3#    feet

Emma Aldridge · Answer

undefined To maximize the volume of the resulting cardboard box, the dimensions of the box will be such that the squares cut out of each corner have equal side lengths. Let's denote this side length as $ x $ feet.

Given that the original square cardboard has side length of 20 feet, after cutting squares of side length $ x $ feet from each corner, the dimensions of the resulting box will be $ (20 - 2x) $ feet by $ (20 - 2x) $ feet by $ x $ feet.

Therefore, the dimensions of the resulting cardboard box will be $ (20 - 2x) $ feet by $ (20 - 2x) $ feet by $ x $ feet.