Minimizing Surface Area of a Box

This video tutorial demonstrates how to minimize the surface area of an open-top box with a fixed volume of 32 cubic feet. The process involves deriving the surface area equation, simplifying it to a function of two variables, and using calculus to find the dimensions that minimize the surface area. The tutorial concludes by verifying the solution through the second partial derivative test, confirming that the dimensions found indeed minimize the surface area.

What is the volume of the box that needs its surface area minimized?

• 128 cubic feet

• 32 cubic feet

• 16 cubic feet

• 64 cubic feet

How many faces does the box have, considering it has no top?

• 4 faces

• 5 faces

• 6 faces

• 7 faces

Which equation is used to express the surface area of the box?

• S = L x W x H

• S = LW + 2HW + 2LH

• S = 2(LW + LH + WH)

• S = L + W + H

What substitution is made to reduce the surface area function to two variables?

• L = H x W / 32

• H = 32 / (L x W)

• W = 32 / (L x H)

• L = 32 / (H x W)

What is the first step in finding the critical points of the surface area function?

• Substitute the dimensions into the surface area equation

• Find the first order partial derivatives

• Solve the volume equation

• Calculate the second order partial derivatives

What is the value of H when the surface area is minimized?

• 3

• 2

• 1

• 4

What is the value of W when the surface area is minimized?

• 2

• 3

• 4

• 5

What is the value of L when the surface area is minimized?

• 5

• 4

• 3

• 2

What is the final minimized surface area of the box?

• 32 square feet

• 56 square feet

• 40 square feet

• 48 square feet

What test is used to verify the relative minimum of the surface area?

• Volume test

• First order partial test

• Surface area test

• Second order partial test