Mathematics, 16.08.2020 01:01 PatienceJoy

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.

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Answer from: ani61

The cost of the side material is $279.50, the cost of the base material is $139.75 and the total cost is $419.25 and this can be determined by using the arithmetic operations.

Given :

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.

Let the width of the rectangular storage container be 'x' then according to the given data the length is '2x' and let the height of the container be 'h'.

The base area of the rectangular storage container is given by:

Base Area = 2x^2 ----- (1)

The side area of the rectangular storage container is given by:

Side Area = 2(2x)h + 2xh

= 6xh ---- (2)

Now, the volume of the rectangular storage container is given by:

Volume = lbh

Put the values of known terms in the above equation.

24 = 2x^2h

$h = \dfrac{12}{x^2}$

Now, the cost is given by:

$\rm C(x)=13(2x^2)+9(6xh)$

$\rm C(x)= 26x^2 + \dfrac{648}{x}$

Now, to minimize C(x) differentiate the C(x) with respect to x.

$\rm C'(x)=52x -\dfrac{648}{x^2}$ ---- (3)

Now, equate the above equation to zero.

$0 = 52x =\dfrac{648}{x^2}$

$x = \sqrt[3]{\dfrac{648}{52}}$

Now, differentiate equation (3) with respect to x.

$\rm C"(x) = 52+\dfrac{1296}{x^3}$

Now, put the value of x in the above equation.

$\rm C"(x) = 52+\dfrac{1296}{\dfrac{648}{52}}$

$\rm C"(x) = 52+104$

$\rm C"(x) = 156 0$

Therefore, 'x' is the point of minima.

Now, the cost of the base material is:

$= 13\times 2 \times (\dfrac{648}{52})^\frac{2}{3}$

= $139.75

Now, the cost of the side material is:

$=\dfrac{648}{\sqrt[3]{\dfrac{648}{52}} }$

= $279.50

Therefore, the total cost is given by:

= 139.75 + 279.50

= $419.25

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Answer from: joannachavez12345

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

Answer from: Quest

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Answer from: Quest

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