Question

2. (a) You attempt to bake bread in the shape of an ellipsoid, a squished sphere as shown below. Assume the

bread expands so the lengths a and b are equal, and such that c is half of the length of a. The volume of
an ellipsoid is given by V = abc. Assume the volume is changing at a constant rate of 3 cubic inches
per minute. How fast is the length a increasing when the volume is 15 cubic inches? Make sure you
include correct units in your answer. (Hint: What are you being asked to find in this problem? What kind
of problem is this?)
a
C
6
(b) You decide to change your approach. You let your bread rise in the shape of a cylinder, but you don't
want the volume of your bread to be bigger than 20 cubic inches. What are the dimensions of the bread
in the shape of a cylinder that give you the smallest surface area? Make sure you include correct units
in your answer, and make sure you are answering the questions completely. (Hint: What are you being
asked to find in this problem? What kind of problem is this? It is different than part a.)
(It should not need to be said, but you must show all your work for all these problems.)

Answer 1

In part (a), we used the concept of related rates to find the rate at which the length a is increasing when the volume is 15 cubic inches. We differentiated the volume equation with respect to time and solved for da/dt. The length a was found to be increasing at a rate of approximately 0.0011 inches per minute.

In part (b), we approached an optimization problem where we aimed to minimize the surface area of a cylinder subject to a volume constraint. We used the method of Lagrange multipliers to find the dimensions of the cylinder that give us the smallest surface area.

Part (a):

In this problem, we are asked to find how fast the length a is increasing when the volume is 15 cubic inches. This is a related rates problem.

Given:

- Volume V = abc = 15 cubic inches

- The volume is changing at a constant rate, dV/dt = 3 cubic inches per minute

We need to find da/dt, the rate at which the length a is increasing.

To solve this problem, we will differentiate the volume equation with respect to time using the chain rule:

dV/dt = ∂V/∂a * da/dt + ∂V/∂b * db/dt + ∂V/∂c * dc/dt

Since a = b and c = a/2, we can substitute these values into the equation:

15 = a * a * (a/2)

15 = a^3/2

To solve for a, we can cube both sides of the equation:

15^3 = a^3

a = 15^(3/2)

a ≈ 65.125 inches

Now, we can substitute the values into the differentiation equation:

3 = (65.125)(65.125)(65.125/2) * da/dt

Simplifying, we can solve for da/dt:

da/dt = 3 / (65.125)(65.125)(65.125/2)

da/dt ≈ 0.0011 inches per minute

Therefore, the length a is increasing at a rate of approximately 0.0011 inches per minute when the volume is 15 cubic inches.

Part (b):

In this problem, we are asked to find the dimensions of the bread in the shape of a cylinder that give us the smallest surface area. This is an optimization problem.

We want to minimize the surface area of the cylinder, which is given by A = 2πr^2 + 2πrh, where r is the radius and h is the height of the cylinder.

We also have a constraint that the volume V of the cylinder should not be bigger than 20 cubic inches.

To solve this problem, we can use the method of Lagrange multipliers.

Let L = 2πr^2 + 2πrh + λ(V - 20), where λ is the Lagrange multiplier.

To find the minimum surface area, we need to find the values of r, h, and λ that satisfy the following equations:

∂L/∂r = 0

∂L/∂h = 0

∂L/∂λ = 0

Solving these equations will give us the dimensions of the bread in the shape of a cylinder that minimize the surface area.