A closed rectangular box is to be constructed with a base as long as its wide suppose that the total surface area is 27square feet find the dimension of the box that maximize the volume?

Answer 1

Since the base is equal to the length, then the two parts of the box, up and down are two squares, let's say with length side x. So the other four parts may be rectangles with length x, and wide y, (where y is also the height of the box). Since the surface area is 27 ft^2, we have: Surface Area = 27 = 2x^2 + 4xy Solve for y y = (27 - 2x^2)/4x

Thus, we have:

l = x

w = x

h = (27 - 2x^2)/4x

V = lwh

V= (x)(x)[27 - 2x^2)/4x] Simplify and multiply:

V = (27x - 2x^3)/4

V = (1/4)(27x - 2x^3) Take the derivative: V'= (1/4)(27 - 6x^2) Find the critical values by setting the derivative equal to zero:

0 = (1/4)(27 - 6x^2) multiply by 0 both sides:

0 = 27 - 6x^2 add 6x^2 to both sides:

6x^2 = 27 divide by 6 to both sides:

x^2 = 27/6

x^2 = 9/2 Square both sides:

x = √(9/2)

x = (3√2)/2

y = (27 - 2x^2)/4x

y = 27/4x - (2x^2)/4x

y = 27/4x - x/2 substitute (3√2)/2 for x:

y = 27/][4(3√2)/2)] - (3√2)/2)/2

y = (9√2)/4 - (3√2)/4

y = (6√2)/4

y = (3√2)/2 As we see the box is a cube.

V = side^3

V = [(3√2)/2]^3 V = (27√2)/4

So, V = (27√2)/4 ft^3 when x = (3√2)/2. Thus, we can maximize the volume of this box if l = w = h = (3√2)/2 ft.