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.

Answers: 2

Show answers

Answers

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

For more information, refer to the link given below:

link

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

answer: was that a question or answer?

step-by-step explanation:

Answer from: Quest

part a

answer: a reasonable domain is $0 \le n \le 10$ where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).