Question

a container with a square bottom, rectangular sides and no top is to e constructed to have a volume of 5 m^3. material for the bottom costs $10 per square meter and material for the sides costs $8 per square meter. find the dimensions of the least expensive container

Answer 1

1.26 m , 3.15 m

Explanation:

Let the side of the square base is y.

Height of the container is h.

Area of base = y x y = y²

Area of side walls = 4 x y x h = 4yh

Volume of the container, V = y²h

According to the question, the volume of the container is 5 m³

So, 5 = y²h

h = 5 / y² .... (1 )

Cost of bottom, C1 = 10 y²

Cost of side walls, C2 = 8 yh

Total cost, C = 10 y² + 8yh

Substitute the value of h from equation (1)

C = 10 y² + 8 y x 5 / y²

C = 10y² + 40 / y

Differentiate it with respect to y

dC/dy = 20 y - 40/y²

For maxima and minima, dC/dy = 0

20 y = 40/y²

y³ = 2

y = 1.26 m

So, h = 5 / (1.26 x 1.26) = 3.15

Thus, the side of base is 1.26 m and height is 3.15 m to minimize the cost.

Answer 2

`2m* 2m* (5)/(4)m`

Explanation:

We are given that a container with square bottom.

Let side of square=x

Height of container=h

Volume of container=
`5m^3`

Cost of 1 square meter of material for the bottom=$10

Cost of 1 square meter of material for side=$8

We have to find the dimension of least expensive container.

Volume of container=
`x^2h`

`x^2h=5`

`h=(5)/(x^2)`

Surface area of container=
`x^2+2(x+x)h=x^2+4xh`

Cost=
`C(x)=10x^2+8(4xh)=10x^2+32x((5)/(x^2))=10x^2+(160)/(x)`

`C(x)=10x^2+(160)/(x)`

Differentiate w.r.t x

`(dC)/(dx)=20x-(160)/(x^2)`

`(dC)/(dx)=0`

`20x-(160)/(x^2)=0`

`(160)/(x^2)=20x`

`x^3=(160)/(20)=8`

`x=\sqrt[3]{8}=2`

Because side of container is always positive.

Again differentiate w.r.t x

`(d^2C)/(dx^2)=20+(320)/(x^3)`

Substitute x=2

`(d^2C)/(dx^2)=20+(320)/(2^3)=60>0`

Hence, the cost is minimum at x=2

Substitute the value of x

`h=(5)/(2^2)=(5)/(4)`m

Hence, the dimensions of the least expensive container are

`2m* 2m* (5)/(4)m`