Question

A rectangular storage container with an open top is to have a volume of 24 cubic meters. The length of its base is twice the width. Material for the base costs 13 dollars per square meter. Material for the sides costs 9 dollars per square meter. Find the cost of materials for the cheapest such container.

Answer 1

Answer:

419.25

Step-by-step explanation:

The calculation of the cost of materials for the cheapest such container is shown below:-

We assume

Width = x

Length = 2x

Height = h

where, length = $2 \times width$

Base area = lb

= $2x^2$

Side area = 2lh + 2bh

= 2(2x)h + 2(x)h

= 4xh + 2xh

Volume = 24 which is lbh = 24

$h = \frac{24}{2x^2} \\\\ h = \frac{12}{x^2}$

Now, cost is

$= 13(2x^2) + 9(4xh + 2xh)\\\\ = 13(2x^2) + 9(4x + 2x)\times \frac{12}{x^2} \\\\ = 26x^2 + \frac{648}{x}$

now we have to minimize C(x)

So, we need to compute the C'(x)

$= 52x - \frac{648}{x^2}$

C"(x)  $= 52x - \frac{1,296}{x^3}$

now for the critical points, we will solve the equation C'(x) = 0

$= 52x - \frac{648}{x^2} = 0\\\\ x = \frac{648}{52}^{\frac{1}{3}}$

$C" = ((\frac{648}{52} ^{\frac{1}{3} } = 52 + \frac{1296}{(\frac{648}{52})^\frac{1}{3} )^3}\\\\ = 52 + \frac{1296}{\frac{648}{52} } >0$

So, x is a point of minima that is

= $(\frac{648}{52} )^\frac{1}{3}$

Now, Base material cost is

$= 13(2x^2)\\\\ = 26(\frac{648}{52} )^\frac{2}{3}$

= 139.75

Side material cost is

$= \frac{648}{x} \\\\ = \frac{648}{(\frac{648}{52})^\frac{1}{3} }$

= 279.50

and finally

Total cost is

= 139.75 + 279.50

= 419.25