Gloria would like to construct a box with volume of exactly 45ft^3 using only metal and wood. The metal costs $15/ft^2 and the wood costs $6/ft^2. If the wood is to go on the sides, the metal is to go on the top and bottom, and if the length of the base is to be 3 times the width of the base, find the dimensions of the box (Length, Width, Height) that will minimize the cost of construction. Round your answer to the nearest four decimal places.
For this problem, first draw a 3D rectangle and label the width as "x", the length as "3x" and the height as "y". This will help you visualize the area formulas presented.
Area Formulas:
Area of top and base= 2(3x)(x)= 6x^2
*This is multiplied by 2 because we have a top and base*
Area of side= xy
Area of 4 sides=8xy
Volume Formula:
Volume of rectangle=LWH
But in this problem we have L=3W so....
Volume (V)=(3W)(W)(H)= 3HW^2
Now substitute y for H and x for W....
Volume (V)=3yx^2
Desired volume is given in the problem as 45 ft^3.....
45=3yx^2
Now we want to relate the Volume formula with the Total Cost formula and we want to only have one variable to solve for. So, solve for Y using the volume formula we just found....
y=45/(3x^2)
Cost Formulas:
Cost of top and base= $15(6x^2)= 90x^2 dollars
Cost of sides= $6(8xy)= 48xy dollars
Total Cost (C)= 90x^2 + 48xy
Substitute y into the total cost formula:
Total Cost (C)= 90x^2 + 48x(45/(3x^2))
C=90x^2 + 2160x/(3x^2))
C=90x^2 +(720/x)
Now take the derivative of the Total Cost formula to achieve our minimum value:
C'=180x-720x^(-2)
Which can also be written as....
C'=180x-(720/x^2)
Now we want to find the critical numbers of the formula, so get a common denominator of the C' formula and set the numerator =0....
*Note, we don't need to set the denominator =0 because that would be where the function does not exist and we are only interested in values that exist for this function.*
C'= (180x^(3)-720)/(x^2)
0=180x^(3)-720
x^3=4
x=+- 4^(1/3), but we are only concerned with positive values...
x=4^(1/3)
x~1.5874
We are almost done!
Length=3x
Length=3(1.5874)
Width=x
Width=1.5874
Height=y
y=45/(3x^2)
*Plug in 1.5874 for x to get y*
Why did the box go to therapy? Because it felt boxed in! Now let's get cracking on solving this problem.
Let's assume the width of the base is x ft. Therefore, the length of the base is 3x ft, and the height of the box is 45ft^3 / (3x * x) ft.
The surface area of the sides (constructed with wood) is 2(3x)(45ft^3 / (3x * x)) ft^2. Simplifying this, we get 90ft^2 / x.
The surface area of the top and bottom (constructed with metal) is 2(3x * x) ft^2, which is 6x^2 ft^2.
The total cost of constructing the box is the cost of the wood plus the cost of the metal. The cost of the wood is (90ft^2 / x) * $6/ft^2, and the cost of the metal is (6x^2) * $15/ft^2.
Adding these costs together, we have a total cost function: C(x) = (90/x) * 6 + 90x * 15.
To minimize the cost, we need to find the value of x that minimizes C(x).
Now, let's go gather the dimensions of the box! Surely this will be a "box-office" hit.
To find the minimum cost, we need to find the value of x that minimizes C(x). We can do this by taking the derivative of C(x), setting it equal to zero, and solving for x.
Differentiating C(x) with respect to x, we get C'(x) = (-540/x^2) + 90 * 15 * 2x.
Setting C'(x) equal to zero and simplifying, we have:
0 = (-540/x^2) + 2700x.
Simplifying further, we get:
540x = x^3 * 2700.
Dividing both sides by 540 and canceling out an x, we have:
x^2 = 5.
Taking the square root of both sides, we get:
x = √5.
Since the width cannot be negative, we will discard the negative value.
Therefore, the width of the base is approximately 2.2361 ft.
The length of the base is 3 times the width, so it is approximately 6.7082 ft.
The height of the box is 45ft^3 / (3x * x) = 4.1503 ft.
Thus, the dimensions of the box that minimize the cost of construction (rounded to four decimal places) are: Length = 6.7082 ft, Width = 2.2361 ft, and Height = 4.1503 ft.
Ah, the box has finally stolen the "show" with its perfect dimensions!
To solve this problem, we need to set up an equation for the cost of construction in terms of the dimensions of the box. Then, we can find the dimensions that minimize the cost by minimizing this equation.
Let's assume the length of the base of the box is x ft and the width of the base is y ft. Since the length of the base is three times the width, we have x = 3y.
The height of the box can be represented as h ft.
First, let's calculate the volume of the box:
Volume = Length x Width x Height
45 = 3y * y * h
45 = 3y^2h
Next, let's find the equation for the cost of construction:
Cost = (Area of metal * Cost of metal) + (Area of wood * Cost of wood)
The area of the top and bottom (which will be made of metal) is:
2 * (Length x Width) = 2 * (3y * y) = 6y^2
The area of the sides (which will be made of wood) is:
2 * (Length x Height) + 2 * (Width x Height) = 2 * (3y * h) + 2 * (y * h) = 6yh + 2yh = 8yh
Now, we can substitute these values into the cost formula:
Cost = 6y^2 * (Price of metal) + 8yh * (Price of wood)
Given that the metal costs $15/ft^2 and the wood costs $6/ft^2, we can substitute these values into the cost equation:
Cost = 6y^2 * 15 + 8yh * 6
Cost = 90y^2 + 48yh
Now, we can substitute the value of h from the volume equation into the cost equation to eliminate one variable:
Cost = 90y^2 + 48(45/(3y^2))
Cost = 90y^2 + 720/y^2
To minimize the cost, we can take the derivative of the cost equation with respect to y and set it to zero:
d(Cost)/dy = 180y - 1440/y^3 = 0
Solving this equation for y, we find:
180y^4 - 1440 = 0
y^4 = 1440/180
y^4 = 8
y = (8)^(1/4) ≈ 1.6818 (rounded to four decimal places)
Now, we can substitute this value of y back into the volume equation to find x:
x = 3y
x = 3(1.6818) ≈ 5.0454 (rounded to four decimal places)
Lastly, substitute the values of x and y into the volume equation to find h:
45 = 3(1.6818)^2h
h ≈ 3.3437 (rounded to four decimal places)
Therefore, the dimensions of the box that will minimize the cost of construction are approximately:
Length ≈ 5.0454 ft
Width ≈ 1.6818 ft
Height ≈ 3.3437 ft
This guy's explanation is weird because he uses L for a length and w for practically the same length. They can both be labeled x's, making the problem less of an eye sore, and h can be labeled y. Your understanding should skyrocket considerably after this redefinition
L = 3 w
metal area = 2 L w = 6 w^2
wood area = 2 L h + 2 w h
volume = 3 w^2 h = 45 so h =15/w^2
cost = 15*6 w^2 + 6*2 w h
c = 90 w^2 + 12 w h
or
c = 90 w^2 + 12 w (15/w^2)
c = 90 w^2 + 180/w
at max or min dc/dw = 0
0 = 180 w - 180/w^2
1/w^2 = w
w = 1^(1/3) = 1
then L = 3
then h = 15
LOL - lots more cheap wood than expensive metal but check algebra