Question

You have been asked to design a rectangular box with a square base and an open top. The volume of the box must be 1715cm3. Determine the dimensions of the bin that will minimize the surface area, where x is the length of each side of the base and y is the height of the box.

Answer 1

The dimensions of the bin that will minimize the surface area are:

Base side (x): 7.5cm

Height (y): 36.13cm

Let's express the surface area and volume of the box in terms of x and y:

Surface area (A):

A = 2x^2 + 4xy

Volume (V)V = x^2y

We are given that the volume of the box must be 1715cm3:

1715cm3 = x^2y

Solving for y, we get:

y = 1715/x^2

Now, we want to minimize the surface area (A) subject to the constraint that the volume (V) is 1715cm3. We can use Lagrange multipliers to solve this optimization problem.

L(x, y, λ) = A - λ(V - 1715)

L(x, y, λ) = 2x^2 + 4xy - λ(x^2y - 1715)

Taking partial derivatives of L with respect to x, y, and λ, we get:

∂L/∂x = 4x + 4y - 2λxy = 0

∂L/∂y = 4x - λx^2 = 0

∂L/∂λ = -x^2y + 1715 = 0

Substituting V = x^2y into the third partial derivative, we get:

∂L/∂λ = -V + 1715 = 0

Solving these equations simultaneously, we get:

x = 7.5

y = 36.13

Answer 2

x=15.08 cm

y=7.54 cm

Explanation:

We are given that

Volume of box=1715 cubic cm

Side length of base=x

l=b=x

Height of box=h=y

Volume of box=lbh=x^2y

`1715=x^2y`

`y=(1715)/(x^2)`

Surface area of box=Area of bottom+area of four faces=
`x^2+4xy`

`S=x^2+4x((1715)/(x^2)=x^2+(6860)/(x)`

Differentiate w.r.t x

`S'(x)=2x-(6860)/(x^2)`

`S'(x)=0`

`2x-(6860)/(x^2)=0`

`2x=(6860)/(x^2)`

`x^3=(6860)/(2)=3430`

`x=(3430)^{(1)/(3))=15.08`

Again differentiate w.r.t x

`S''(x)=2+(13720)/(x^3)`

Substitute x=15.08

`S''(x)=2+(13720)/((15.08)^3)>0`

Hence, the surface area is minimum at x=15.08 cm

`y=(1715)/((15.08)^2)=7.54 cm`