The top and bottom of a can must be made of copper that costs two cents per square inch and the curved side is to be made of aluminum that costs one cent per square inch. Find the radius and height of the can that will maximize the volume of the can where the total cost of the can is 300pi cents.

Cost=.01*area of Al + .02*area of ends

300=.01*PI*radius*2*height+.01*PI*2*radius^2

Use that to find height as a functionof radius.

Now
Volume=PI*radius^2 * Height
set dVolume/drad =0 and solve for radius, then go back and solve for height.

Well, well, well, let's get calculating. So we have a can with copper top and bottom, and an aluminum curved side. The copper costs 2 cents per square inch, while aluminum costs 1 cent per square inch. We want to maximize the volume of the can while keeping the total cost at 300π cents. Let's do this!

Let's start by breaking down the cost of the can. We know that the top and bottom are made of copper, while the curved side is made of aluminum. The formula for the cost of the can is given by:

Cost = copper cost + aluminum cost.

The copper cost is 2 cents per square inch, and the aluminum cost is 1 cent per square inch. So we have:

Cost = 2cπr² + 1cπrh, where c is the constant cost per square inch factor, r is the radius of the can, and h is the height of the can.

We want the total cost to be 300π cents, so we have:

300π = 2cπr² + 1cπrh.

Now, let's move on to maximizing the volume. The formula for the volume of a cylinder is given by:

Volume = πr²h.

Since we want to maximize the volume, we are essentially looking for the maximum value of the function Volume = πr²h.

So, we have two equations now:

300π = 2cπr² + 1cπrh, and
Volume = πr²h.

Now, let me put on my mathematical clown hat and juggle these equations for you. We can solve the first equation for h:

h = (300π - 2cπr²) / (cπr).

Now, let's substitute this value of h into the second equation:

Volume = πr²h = πr²[(300π - 2cπr²) / (cπr)].

Simplifying, we have:

Volume = (300πr - 2cπr³) / c.

Now, to find the values of r and h that maximize this volume, we take the derivative of the volume equation with respect to r, set it equal to zero, and solve for r. But be warned, this could get a bit derivative!

(d/dx)(Volume) = (d/dx)[(300πr - 2cπr³) / c] = 0.

Simplifying even more, we have:

(d/dx)(Volume) = (300π - 6cπr²) / c = 0.

Solving for r, we get:

300π - 6cπr² = 0,
6cπr² = 300π,
r² = 300/6,
r² = 50,
r = √50.

Finally, to find the height (h), we substitute this value of r into the equation for h we derived earlier:

h = (300π - 2cπ(√50)²) / (cπ√50).

Simplifying, we get:

h = (300π - 200cπ) / (c√50).

Now, let's put on the clown wig and plug in those values of r and h into the equations to get the maximum volume of the can. That's worth clowning around for, right?

To find the radius and height that will maximize the volume of the can while considering the cost, we need to set up an equation for the total cost and an equation for the volume of the can.

Let's start with the equation for the total cost:

Cost = (Cost of copper top and bottom) + (Cost of aluminum side)

Since the copper top and bottom cost 2 cents per square inch, and the aluminum side costs 1 cent per square inch, we can write the cost equation as:

Cost = 2πr^2 + (2πrh)(1)

Where r is the radius of the can, and h is the height of the can.

Now, let's set up the equation for the volume of the can:

Volume = (Volume of cylinder - Volume of top and bottom)

The volume of a cylinder is given by:

Volume of cylinder = πr^2h

The volume of the top and bottom is given by:

Volume of top and bottom = 2(πr^2)

Substituting the above equations, the volume equation becomes:

Volume = πr^2h - 2(πr^2)(2)

Now, we are given that the total cost of the can is 300π cents. Substituting this into the cost equation, we have:

300π = 2πr^2 + 2πrh

Simplifying this equation, we get:

150 = r^2 + rh

Now, we have two equations:
1. Volume = πr^2h - 2(πr^2)
2. 150 = r^2 + rh

To maximize the volume, we can solve equation (2) for h:

h = (150 - r^2)/r

Substituting this value of h into equation (1), we have:

Volume = πr^2((150 - r^2)/r) - 2(πr^2)

Simplifying this equation, we get:

Volume = πr^2(150 - r^2)/r - 2πr^2

Now, we can find the derivative of the volume equation with respect to r:

dV/dr = (2πr(150 - r^2)/r^2) - 4πr

Setting the derivative equal to zero to find the critical points:

0 = (2πr(150 - r^2)/r^2) - 4πr

Simplifying this equation, we get:

0 = 300 - 3r^2 - 2r^2

0 = 300 - 5r^2

5r^2 = 300

r^2 = 60

r ≈ 7.75 (taking the positive value)

Now, we can substitute this value of r into equation (2) to find the corresponding value of h:

150 = 7.75^2 + 7.75h

150 = 60 + 7.75h

7.75h = 90

h ≈ 11.6

Therefore, the radius of the can that will maximize the volume is approximately 7.75 inches, and the height is approximately 11.6 inches.

To solve this problem, we need to find the dimensions of the can that will maximize its volume while still keeping the total cost within the given constraint. Let's break it down step by step.

1. Define the variables:
- Let's use "r" to represent the radius of the top and bottom of the can.
- Let's use "h" to represent the height of the can.

2. Determine the cost of the top and bottom:
- The top and bottom of the can are made of copper, which costs 2 cents per square inch.
- Each of the top and bottom is a circle, so the cost of the top and bottom can be calculated as: Cost_top_and_bottom = 2 * π * r^2 * 2 cents.

3. Determine the cost of the curved side:
- The curved side of the can is made of aluminum, which costs 1 cent per square inch.
- The curved side is a rectangle that is formed by rolling a rectangular strip with a width equal to the circumference of the circular top/bottom.
- The circumference of a circle is given by: Circumference = 2 * π * r.
- So, the cost of the curved side can be calculated as: Cost_curved_side = Circumference * h * 1 cent = (2 * π * r) * h * 1 cent.

4. Calculate the total cost of the can:
- The total cost of the can is given as 300π cents.
- So, the total cost can be calculated as: Total_cost = Cost_top_and_bottom + Cost_curved_side.

5. Express the volume of the can in terms of r and h:
- The volume of the can is given by volume = π * r^2 * h.

6. Derive an equation for one variable in terms of the other:
- Rearrange the total cost equation to solve for h in terms of r: h = (300π - 4πr^2) / (2πr + 2πr)
- Simplify: h = (300 - 4r^2) / (4r) = (75 - r^2) / r.

7. Express the volume equation in terms of a single variable:
- Substitute the equation derived in step 6 into the volume equation: volume = π * r^2 * ((75 - r^2) / r).

8. Maximize the volume:
- Find the derivative of the volume equation with respect to r: dv/dr = (d/dx) (π * r^2 * ((75 - r^2) / r)).
- Simplify the derivative: dv/dr = -π * (3r^2 - 75) / r^2.
- Set the derivative equal to zero to find the critical points: -π * (3r^2 - 75) / r^2 = 0.
- Solve for r: 3r^2 - 75 = 0.
- Simplify and solve for r: r^2 = 25.
- Take the positive square root to find the value of r: r = 5.

9. Find the corresponding value of h:
- Substitute the value of r into the equation derived in step 6: h = (75 - (5^2)) / 5.
- Simplify: h = 10.

So, to maximize the volume of the can while keeping the total cost within the given constraint, the radius should be 5 units and the height should be 10 units.