im really stuck on this problem, any help?

Question

a container with rectanglar base, rectangular sides, and no top is to have a volume of 16m^3. The width of the base is to be 2m. When cut to size, material costs $27 per square meter for the base and $9 per square meter for the sides. What is the cost of the LEAST expensive container?

Answer 1

The key to questions like this is the proper setup. Eventually, you want to get the cost into a function of one variable that you can differentiate, but first start with all the variables you need to describe the problem. Let a=width, b=length, c=height of the box. Then you know a=2 and abc=16, so bc=8.

Now you want to minimize the cost which is:
27ab + 9(2ac + 2bc) = 54b + 36c + 144

Now since bc = 8, you can pick one either b or c and plug in for it into the cost equation in term of the other. Let substitute in for c, with 8/b. Then we get that the cost of the box is: 54b + 288/b + 144.

Differentiating with respect to b, we get:
54 - 288/b^2 = 0 so b = sqrt(288/54) = sqrt(16/3), so since bc = 8, c=sqrt(12). Thus the dimensions of the box (unless, I made a silly arithmetic/algebra mistake) are width = 2, length = sqrt(16/3), and height = sqrt(12).

Hope that helps!

Answer 2

The target is the total cost of materials which are the function of areas and the unit costs of materials..

Let
L = the length
W = the width
H = the height
B = the base area
S = the side area
C = the total cost


B = LW = 2L
S = 2(L+W)H = 2(L+2)H

From LWH = 16 and W = 2, we have
2LH = 16
H = 8/L......(1)

Since we know the unit costs for the base and the side, the total cost can be written as
C = 27B + 9S

Plug in B = 2L and S = 2(L+2)H into above equation,
C = 27(2L) + 9[2(L+2)H] ......(2)

Plug in (1) into (2),
C = 54L + 18(L+2)(8/L)......(3)

Differentiate (3) with respect to L,
C' = 54 - 288/L^2......(4)

Solve C' = 0 for L,
L = 2.3094 m......(5)

H = 8/L = 3.4641 m......(6)

Plugging in (5) and (6) into (2) gives the least cost,
C = $393.42

Answer 3

First, try to think about what the goal is. The goal is to find the dimensions of the container that would minimize cost. Therefore, you'll have to find the minimum of the cost function by taking its derivative.

How do you get the cost function? Write out all the equations and information you know.

Make an equation for the volume of the container. Let's say H is the height, W is the width, and L is the length. You know already that the width is 2.

WHL=16. (* is multiplication)
2HL=16 (put in 2 for the width)
HL=8

Make another equation for the surface area.

(Area of the sides) + (Area of the base)=surface area
(2HL+2HW)+(LW)=surface area (S)
(2HL+4H)+(2L)=S (put 2 in for the width)
(2HL+4H)+(2L)=S

Now, you have two unknowns (H and L) and two equations. Combine them to get one variable you can eventually solve for.

HL=8
L=8/H

2H(8/H)+4H+2(8/H)=S
16+4H+16/H=S

Remember the first two terms are still the total area of the sides and the last is still the total area of the base.

Now, you can turn this into the cost function by multiplying the cost of the sides with the cooresponding area of the sides and the cost of the base with the cooresponding area of the base.

9(16+4H)+27(16/H)=Cost
144+36H+432/H=Cost
144+36H+432*H^(-1)=Cost

Now take the derivative of the cost function:

Cost’=36-432*H(^-2)

Find where Cost’=0 to find the minimum point of the cost function.

0=36-432*H(^-2)

432*H^(-2)=36

12=H^2

sqrt(12)=H

Now you can plug this H value back into the original Cost function to get the minimum cost.

Hope this helps!