I know that this problem is an optimization problem, and have been going through that section in my textbook. I feel like I understand the examples but am unsure how to begin with this problem or if I am on the right track with my calculations. Any direction or hints are appreciated!
Problem: You work for a fancy packaging company. Your client wants you to produce containers that will hold a very valuable mineral in powder form. The containers will be made of a material that costs 10/cm2 and
you can spend $1000 per container. Furthermore, the container needs to be in the shape of a pyramid with square base when folded up. Your task is to design a container (find the length of the base and height of the pyramid) that has the maximum possible volume, and costs exactly $1000 to produce.
a. What facts/formulas will you need? (I said we would need to know how to find the first and second derivative, the formula for the surface area of a square and four triangles, the formula for the volume of a pyramid, and the first derivative test for Absolute Extreme Values as well as how to derive Extreme Values)
b. Write the equation that you need to optimize, and also an equation that describes the constraint. (I'm not sure how to write the equation when we use different formulas for area and volume, but I was thinking that it would be a problem where I set the surface area equal to the volume, and the constraint would be the cost has to be equal to $1000)
C. Show that the equation you need to maximize, as a function of a single variable x (the length of the pyramid's base) is V = (5/3)(sqrt(100x^2-2x^4))
d. Maximize your equation either by using derivatives or by graphing or both. Then write down the dimensions of the pyramid (radius and height) that will lead to the pyramid with maximum volume possible (given your constraint). Indicate also the (now optimal) volume of your pyramid.
If the pyramid has base side 2x and height h, then you know that it must have an area of 100 cm^2. The slant height of each face is √(h^2+x^2) so that means that
(2x)^2 + 4(x√(h^2+x^2)) = 100
h^2 = 25(25-2x^2)/x^2
Now, the volume is
v = 1/3 (2x)^2 h = 4/3 x^2 h = 20/3 x√(25-2x^2)
dv/dx = 20/3 (25-4x^2)/√(25-2x^2)
so max volume is when x = 5/2
That is, when the pyramid has a base with side = 5cm
Note that my solution is a bit different, since I used 2x for the base side to avoid fractions. But the answer is the same. You can massage it as you see fit.
Did you make a reasonable 3D sketch of the pyramid?
I let the base of the pyramid be 2x by 2x, (avoiding some fractions later on)
and let the height of the pyramid be y
So we know the volume = (1/3)(4x^2)(y) , let that one sit for a while
Look at one of the triangles, let its height be h,
then h^2 = x^2 + y^2 , (this is why I let the base be 2x)
h = √(x^2 + y^2)
I will assume the cost is $10/cm^2 or is it 10 cents/cm^2 , this will be a problem
I will assume it is $10
cost of 1 pyramid = 10(4x^2) + (10)(4)(1/2)(2x)(h) = 1000
4x^2 + 4xh = 100
x^2 + xh = 25
h = (25-x^2)/x
Now back to our h = √(x^2 + y^2)
√(x^2 + y^2) = (25-x^2)/x
square both sides:
x^2 + y^2 = (625 - 50x^2 + x^4)/x^2
x^4 + x^2 y^2 = 625 - 50x^2 + x^4
x^2 y^2 = 625 - 50x^2
y = √(625-50x^2)/x
and now back into V = (1/3)(4x^2)(y)
= (1/3)(4x^2)√(625 - 50x^2)/x
= (4/3) x √(625 - 50x^2)
= (4/3)√(625x^2 - 50x^4)
dV/dx = (4/3)(625x^2 - 50x^4)^(-1/2) (1250x - 200x^3) = 0
This can only be true if 1250x - 200x^3 = 0
50x(25 - 4x^2) = 0
x = 0 , which clearly would produce a minimum volume of 0
or
4x^2 = 25
x = 5/2
so the base is 5 by 5
y = √(625-50x^2)/x
= √(625 - 312.5)/2.5 = 7.071
I did not check my calculations, I should have written the solution out first
You better check this mess
To begin solving the problem, let's go step by step:
a. Facts and Formulas:
- We need to find the first and second derivative.
- Formula for the surface area of a square: A = s^2, where s is the length of the side of the square.
- Formula for the surface area of four triangles (pyramid sides): A = 4 * (1/2) * b * h, where b is the base length of the triangle and h is the height of the triangle.
- Formula for the volume of a pyramid: V = (1/3) * base area * height.
- First derivative test for Absolute Extreme Values and how to derive Extreme Values.
b. Equations:
- The equation we need to optimize is the one for volume, V.
- The constraint is the cost, which needs to be equal to $1000.
c. Equations to Maximize:
First, let's write down the equation for volume (V) in terms of a single variable x, the length of the pyramid's base:
To do this, we'll need to start by finding expressions for the base area and height of the pyramid.
Base area (A_base):
Since the base of the pyramid is a square, and its length is x, the base area can be calculated as A_base = x^2.
Height (h):
Using the Pythagorean theorem, we can find the height of the pyramid (h) in terms of x:
The hypotenuse of each triangular side is x because it matches the length of the square base. The height is formed by half of the diagonal, which can be found by dividing x by √2.
So, the height (h) can be calculated as h = (1/2) * (x / √2).
Now, we can substitute the values of A_base (x^2) and height (h) into the volume equation to get:
V = (1/3) * A_base * h
V = (1/3) * x^2 * (1/2) * (x / √2)
Simplifying further, we have:
V = (1/3) * x^3 / √2
However, we also need to consider the constraint, which is the cost:
The cost is equal to the area of the material used in the container. Since the container's shape is a folded pyramid, we need to find the total surface area.
Surface area (A_surface):
To calculate the surface area, we add the areas of the base and four triangular sides:
A_surface = A_base + 4 * (1/2) * (x^2 / √2) * (x / √2)
Simplifying further:
A_surface = x^2 + 4 * (1/2) * (x^3 / 2)
A_surface = x^2 + x^3
Now, we have the volume equation (V) and the surface area equation (A_surface), both in terms of x.
d. Maximize the Equation:
To find the maximum volume, we need to maximize the equation V = (1/3) * x^3 / √2, while keeping the constraint A_surface = x^2 + x^3 equal to $1000.
Differentiating V with respect to x, we have:
dV/dx = (1/3) * (3/x^2) / √2
Simplifying further:
dV/dx = 1 / (x^2√2)
Now, let's set the equation for the surface area equal to $1000:
A_surface = x^2 + x^3 = 1000
To find the values of x that maximize V, we can use the first derivative test for Absolute Extreme Values:
- Set dV/dx = 0 and solve for x.
- Determine whether the critical points found are maximum or minimum points by analyzing the second derivative (if needed).
After finding the critical values of x, substitute them back into the volume equation to find the corresponding dimensions (length and height) and the optimal volume.
Remember to check the maximum or minimum nature of the critical values using the second derivative test.