open rectangular box with square base

Question

An open rectangular box with a square base has a surface area of 300 square inches. Find the dimensions that give the maximum volume. What is the maximum volume?

HINT: The surface area of an open rectangular box with a square base of length x and width x and height h is given by the formula A= X^2 + 4xh and the volume is given by the formula V= x^2 h

Answer 1

This is a very common question type for Calc AB, known as an optimization question. You are given a quantity to maximize or minimize (in this case the volume of the box), and you are given a constraint (in this case, the SA = 300 in^2). When you have a function in only one variable, it is relatively straightforward to take its derivative, set that deriv = 0, and solve to find the critical pt (either a max or a min). The problem with doing that for V = x^2h is that we have TWO variables. So, we use the constraint, and solve for h to get h in terms of x: SA = x^2 + 4xh = 300 4xh = 300 - x^2 h =(300 - x^2) / (4x) Next, substitute that expression in for h in the volume equation: V = x^2 (300 - x^2) / (4x) Simplifying, V = 75x - (1/4)x^3 Take deriv, set = 0, solve: V' = 75 - (3/4)x^2 = 0. x = 10 The dims of the rectangular box of max volume are 10" x 10" x 5". The max volume is 500 cubic inches. You can prove that this critical value, x = 10, yields a MAX for volume by showing that the derivative goes from being + to - there. Thus, the volume function is increasing for x < 10, and decreasing thereafter.

Answer 2

If A = x^2 + 4xh, 4xh = A - x^2, or h = (A - x^2)/4x

Thus, V = x^2h =x^2(A - x^2)/(4x) = Ax/4 - x^3/4

dV/dx= 0

A/4 - 3x^2/4 = 0

3x^2/4 = A/4

x^2 = A/3

x= √(3A)/3

d²V/dx² = -3x/2

This is negative for all positive x, including our solution, so this is a maximum.

As A = 300 in²

x = √(3*300 in²)/3 = 30 in/3 =10 in

h = (A - x^2)/(4x) = (300 - 10^2)/(4*10) = 200/40 = 5in

The maximum volume is V = x²h = (10 in)²(5in) = 500 in³

Answer 3

hb^2 = Volume (using b=x for base side)

A= 4hb + b^2 = 300

4hb = 300-b^2

h = (300-b^2)/4b

V=(300-b^2)/4b x b^2 = (300-b^2)b/4 = 75b -(1/4)b^3

V'(b) = 75-(3/4)b^2 = 0

b^2 = (4/3)75 = 100

b = 10 inches

h = 5 inches