Cylinder and cones (Putnam Exam 1938) Right circular cones of ...

Question

Cylinder and cones (Putnam Exam 1938) Right circular cones of height h and radius r are attached to each end of a right circular cylinder of height h and radius r, forming a double-pointed object. For a given surface area A, what are the dimensions r and h that maximize the volume of the object?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says an engineer is designing a water tank that consists of a cylindrical part flanked by two hemispherical ends. If the total external surface area of the tank is to be 452 m squared, what should be the radius R of the tank to maximize its volume? We give 4 possible choices as our answers. For choice A, R is equal to the square root of the quantity of 226 divided by pi in quantity meters. For choice B, we have R is equal to the square root of the quantity of 113 divided by pi in quantity meters. For choice C we have R is equal to the square root of the quantity of 113 divided by the quantity of two pi in quantity meters. For choice D we have R is equal to the square root of the quantity of 113 divided by the quantity of 3 pi in quantity meters. Now we're asked to determine the radius of our tank that would maximize its volume. So first we need to determine the volume of the tank. So we'll call that volume V. And it's going to be equal to the volume of the cylindrical part. We call that for the volume of a cylinder, we have pi R squad, multiplied by H. Where H is the height of the cylinder, but we'll have to add to this the, the volume of the two hemispherical ends. So, if I have two hemispheres, that means I have a total sphere, and then the volume of a sphere, if you recall, is going to be 4 divided by 3, multiplied by pi cubes, so I'll have to add to the volume of our cylinder. 4 divided by 3, multiplied by pi or cube, which is the volume of the sphere. Now, I ultimately want to take a derivative of our volume with respect to one of these quantities, actually the radius. So what I want to do is rewrite my height in terms of the radius. And I can do that by using our surface area. So, our surface area A of this tank is going to be equal to the surface area of the cylindrical part, and if you recall the surface area of a cylinder, that is going to be 2 pi are multiplied by H, then we'll have to add to that the surface area of the two hemispheres, which is just going to be the surface area of the sphere, which is going to be 4I R squared. Now, the reason why we are looking at the area because we're told that the total surface area of the tank has to be 452 m squared. So that means our area here has to be equal to 452. So what I can do is solve this for H. So, to start off, I also track the 4 pi square from both sides. So I'll have 2 pi R multiplied by H is equal to. 452 minus 4 squad. Now I'll just divide both sides by 2 pi R, and we'll get H is equal to the quantity of 452 minus or r squad in quantity divided by the quantity of 2 pi r. So now will substitute this expression and for H into our equation for our volume. That means our volume is going to be equal to. Pi R squared multiplied by the quantity of 452. Minus or pi are squared in quantity divided by the quantity of 2 pi R in quantity. They'll have plus 4 divided by 3, multiplied by pi or cubed. Now we're just going to simplify this expression. So this is going to be equal to. 226 R. Minus Two pi are cubed. Plus, we're divided by 3. Multiplied by i or cued. And combining the last two terms, this is going to be equal to. 226 R. Minus 2 pi are cubed divided by 3. So now we want to do is maximize this volume with respect to R. So we want to take a derivative of our volume with respect to R and then set it equal to 0 and solve for R. So, first taking the derivative of V with respect to R. This is going to be equal to the derivative with respect to R. Of the quantity of 226 R minus. 2 pi are cubed divided by 3 in quantity. And this is going to be equal to, and here we're just going to apply our power rule and constant multiple rule. Take this derivative. So this is going to be equal to 226. Minus 2 pi are squared. So now what we need to do is set this equal to 0 and solve it for R. We'll have 226. Minus 2 pi, multiplied by R2 is equal to 0. So to solve this for, first thing I want to do is add 2 pi R squad to both sides. And so that means I will have. 226 equal to 2. Are squared. So now I'll divide both sides by 2 pi. I will get. 113 divided by pi equal to R2. And so now I just need to take the square root of both sides, and so that means that R is going to be equal to the square root of the quantity of 113 divided by pi in quantity. And here we've taken the positive square root, since R has to be a positive quantity, we can't have a negative radius. So, now we need to check this, make sure that this is actually a maximum. And to do this, we're gonna need to take another derivative of our volume with respect to R. So we're gonna calculate the second derivative. We'll have the second derivative of V with respect to R. And this is going to be equal to the derivative with respect to R. Of the quantity of 226 minus 2 pi or squared. In quantity And so taking this derivative, this is going to be equal to minus or pi R, which is actually going to always be negative. So it's always going to be less than 0, since R has to be a positive quantity, and we have the minus sign out front, making this overall a negative quantity. And we call that if you have a negative second derivative, that means that our function is going to be concave down. And if we're looking at for our value, That is supposed to be on an extrema or a function that is concave down, that means that we actually have a maximum. So our R value that we calculated is the maximum, the the radius that will maximize our volume. So that is the answer that we're looking for. And so if we could compare our our value. To the ones in our answer choices, we see that the correct answer choice corresponds to answer choice B. So I hope this explanation has been useful, and I'll see you in the next video.

Cylinder and cones (Putnam Exam 1938) Right circular cones of height h and radius r are attached to each end of a right circular cylinder of height h and radius r, forming a double-pointed object. For a given surface area A, what are the dimensions r and h that maximize the volume of the object?

Key Concepts

Surface Area and Volume Formulas

Optimization Techniques

Calculus of Functions of Multiple Variables

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