An open-top box having a square base and a surface area of 120 square inches. size make max. volume?L of base

Question

"A manufacturer wants to design an open-top box having a square base and a surface area of 120 square inches. What dimensions will provide a box with maximum volume?
Length of base ?
Width of base?
Height of box ?"

I have no idea where to even begin! HELP!

Answer 1

hmmm..
if i remember it correctly...

Let V=ha^2
let A=a^2+4ah=120

partial dv=a^2+2ah
set dv=0 and we have two equations

a^2+2ah=0 and
a^2+4ah=120
two equations and two unknowns hmmm...and got stock

or we can take the equation for Area and find h

h=(120-a^2)/4a
substitute h in V and we have
V=120a-a^3

V'=120-3a^2

a=sqrt(120/3)=sqrt(40)=6.32
then
h=3.16

Answer 2

First you need to figure an equation for the surface area. Let's let x = the length and width of base (it's a square). Let h represent the height of the box.

We know that we have one side (base) of area x * x, and 4 sides of area x * h. That means we have the equation:

x * x + 4 ( x * h ) = 120.

Let's solve for h:
x² + 4xh = 120
4xh = 120 - x²
h = (120 - x²) / 4x

Okay, now volume will be:
V = x² * h

Substitute in what we determined for h above:
V = x²(120 - x²) / 4x

Canceling an x:
V = x(120 - x²) / 4

Now multiply through:
V = 30x - x^3/4

Now you have an equation that you can take the derivative and set to zero. This should give you the maximum.

dV/dx = 30 - 3x² / 4 = 0

So 3x² / 4 = 30
3x² = 120
x² = 40

That gives two answers (x = +/-sqrt(40) ) but since this is a length, only the positive square root makes sense.

So the volume will be maximized when the base is equal to sqrt(40):
x = sqrt(40)
x = 2 sqrt(10) = 6.324 inches for each base.

And now you can compute the height using the prior equation for h:

h = (120 - x²) / 4x
h = (120 - 40) / 4sqrt(40)
h = 80 / 8 sqrt(10)
h = 10 / sqrt(10)
h = sqrt(10) = 3.162 inches for the height.

A double check confirms that the surface area is indeed 120 sq. inches. The base is 40 square inches, and each side is 20 sq. inches, for a total of 120 square inches.

Your open-top box will have a maximum volume (of 20 cu. inches) when:
Base = 2sqrt(10) = 6.324 inches
Height = sqrt(10) = 3.162 inches

Answer 3

V = L²H

Also L² + 4LH = 120
Thus H = (120 - L²)/(4L)

Thus V = L²(120 - L²)/(4L)
= 30L - (L³)/4

dV/dL = 30 - 3L²/4 = 0 for stationary points
So 3L² = 120
L² = 40
L = ± 2√(10)
Since L > 0 L = 2√(10)

d²V/dL² = -3V/2 < 0 when L = 2√(10). Therefore local maximum at L = 2√(10) in. (You can show minima in these constraints occur either when L = 0 or H = 0 (in both cases V = 0)

H = (120 - L²)/(4L)
= (120 - 40)/(8√(10))
=80/(8√(10))
=√(10) in

Thus L = B = 2√(10) in. and H = √(10) in

Vmax= L²H = 40 * √(10)
=40√(10) in³

Check: Surface Area = L² + 4LH
=40 + 4 *2√(10)*√(10)
= 40 + 80
= 120 in²