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 $20 per square meter. Material for the sides costs $12 per square meter. Find the cost of materials for the cheapest such container. (Round your answer to the nearest cent.) $

Answer 1

Answer:

$ 327.08

Step-by-step explanation:

Let w be the width ( in meters ) of the container,

⇒ Length of the container = 2w,

If h be the height of the container,

So, the volume of the container = length × width × height

= 2w × w × h

= 2w² h

According to the question,

$2w^2h=10$

$\implies h=\frac{10}{2w^2}=\frac{5}{w^2}$

Now, the area of the base = length × width

$=2w^2$

Area of sides = 2 × length × height + 2 × width × height

$=2\times 2w\times h+2\times w\times h$

$=4w\times \frac{5}{w^2}+2w\times \frac{5}{w^2}$

$=\frac{20}{w}+\frac{10}{w}$

$=\frac{30}{w}$

Since, material for the base costs $20 per square meter and material for the sides costs $12 per square meter,

Hence, total cost,

$C(w) = 2w^2\times 20 +\frac{30}{w}\times 12$

$C(w)=40w^2+\frac{360}{w}$

Differentiating with respect to w,

$C'(w) = 80w - \frac{360}{w^2}$

Again differentiating with respect to w,

$C''(w) = 80 +\frac{720}{w^3}$

For maxima or minima,

C'(w) = 0

$80w - \frac{360}{w^2}=0$

$80w^3-360=0$

$80w^3=360$

$\implies w=\sqrt[3]{\frac{360}{80}}=1.65096362445\approx 1.651$

For w = 1.651, C''(w) = positive,

Thus, cost is minimum for width 1.651 meters,

And, the minimum cost = C(1.651) = $40(1.651)^2+\frac{360}{1.651}=\ 327.081706869\approx \ 327.08$