Question

A rectangular storage container with an open top is to have a volume of 10 m3. The length of this base is twice the width. Material for the base costs $15 per square meter. Material for the sides costs $9 per square meter. Find the cost of materials for the cheapest such container. (Round your answer to the nearest cent.) $

Answer 1

Answer:

Cost of material is $245.31

Step-by-step explanation:

Let dimension of box,

• Length, l = x m

• Width, w = y m

• Height, h = z m

The length of this base is twice the width.

∴ x = 2y

Volume of box = 10 m³

∴ xyz = 10

⇒ 2y²z = 10

⇒ y²z = 5

$\Rightarrow z=\dfrac{5}{y^2}\ \ \ ...(i)$

Material for the base costs $15 per square meter.

Total cost of base = 15xy

Total cost of base = 30y²         [∵ x = 2y ]

Material for the sides costs $9 per square meter.

Total cost of side = 9(2xz+2zy)

Total cost of side = 18(xz+yz)

Total cost of material for container = 30y² + 18(xz+yz)

$C(y)=30y^2+18(2y\cdot \dfrac{5}{y^2}+y \dfrac{5}{y^2})$     [From (i)]

$C(y)=30y^2+\dfrac{270}{y}$

Differentiate w.r.t y

$C'(y)=60y-\dfrac{270}{y^2}$

For critical point , C'(y)=0

$60y-\dfrac{270}{y^2}=$

$y=\sqrt[3]{\dfrac{9}{2}}$

$x=2y=2\sqrt[3]{\dfrac{9}{2}}$

$z=\dfrac{5}{y^2}=5\sqrt[3]{\dfrac{4}{81}}$

The minimum cost of container material  at $y=\sqrt[3]{\dfrac{9}{2}}$

$C_{min}=30\sqrt[3]{\dfrac{81}{4}}+270\sqrt[3]{\dfrac{2}{9}}$

$C_{min}=245.31$

Hence, the cheapest cost of material for container is $245.31