For its beef stew, Betty Moore Company uses aluminum containers that have the form of right circular cylinders. Find the radius and height of a container if it has a capacity of 29 in.3 and is constructed using the least amount of metal. (Round your answers to two decimal places.)
I am unsure as to how to set this problem up and solve it. All answers appreciated, thank you.
To solve this problem, we can use the formula for the volume of a cylinder, which is given by:
V = πr^2h
Here, V represents the volume, r represents the radius, and h represents the height of the cylinder. We are given that the capacity of the container is 29 in^3, so we can set up the equation as follows:
29 = πr^2h
To minimize the amount of metal used, we want to minimize the surface area of the cylinder, which is given by:
A = 2πrh + πr^2
We can express h in terms of r using the equation above and then substitute it into the equation for the volume. Let's rearrange the surface area equation to solve for h:
2πrh = A - πr^2
h = (A - πr^2) / (2πr)
Substituting this expression for h into the volume equation:
29 = πr^2 ((A - πr^2) / (2πr))
29 = (A - πr^2) / (2r)
58r = A - πr^2
πr^2 + 58r - A = 0
To find the values of r and h that minimize the amount of metal used, we need to minimize the surface area. We can do this by finding the minimum value of A. To find the value of r and h that minimize A, we can take the derivative of A with respect to r, set it equal to zero, and solve for r.
Let me calculate that for you.
To solve this problem, we can use the formula for the volume of a cylinder, which is given by:
V = πr²h
Where:
- V is the volume of the cylinder
- r is the radius of the base
- h is the height of the cylinder
In this case, we are given that the volume of the cylinder is 29 in³. Since we want to minimize the amount of metal used, we need to minimize the surface area of the cylinder.
The surface area of the cylinder can be calculated by adding the area of the two circular bases and the area of the curved lateral surface:
A = 2πrh + 2πr²
To minimize the surface area, we can differentiate this expression with respect to one of the variables (r or h) and set the derivative equal to zero:
dA/dr = 2πh + 4πr = 0
Solving this equation, we get:
h = -2r
Now, we can substitute this expression for h into the volume equation:
29 = πr²(-2r)
Simplifying this equation, we have:
29 = -2πr³
Divide both sides by -2π:
r³ = -29/(2π)
To solve for r, take the cube root of both sides:
r = ∛(-29/(2π))
Plug in the value of π (approximately 3.14159) and compute the value of r using a calculator:
r ≈ 1.26
Now that we have the value of r, we can substitute it back into the equation for h:
h = -2r ≈ -2(1.26) = -2.52
Since height can't be negative in this context, we take the absolute value:
h ≈ 2.52
Therefore, the radius of the container is approximately 1.26 inches and the height is approximately 2.52 inches.
just like all the other related-rates problems.
v = pi r^2 h = 29
h = 29/(pi r^2)
the metal used is just the surface area
a = 2*pi r^2 + 2 pi r h
= 2pi r^2 + 2pi r * 29/(pi r^2))
= 2pi r^2 + 58/r
da/dr = 4pi r - 58/r^2
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