 In this video we present the solution to question number 14 for practice exam number four for math 1210 And it's an optimization problem Suppose that a can is made to contain a thousand pi cubic centimeters of liquid The reason the pie there is just to make this thing Arithmetically simpler to compute for you, especially if you don't have a scientific calculator with you So before we read the rest of it, let's kind of imagine we have this can a can just means in this situation It's a cylinder The volume of the can is actually volume of the cylinders actually provided to us here So we see the volume is going to equal pi r squared h and by the constraint given this volume is going to have to be 1000 pi well the radius of the circle on top That's what r is and h is then measuring the height of this can right here So suppose that it costs five cents per square centimeter for the metal use on the top and bottom of the can But it costs 10 cents per cubic centimeter Excuse me per square centimeter for the metal on the side So it turns out that that the top and bottom costs a little bit different than the side And we want to find the exact cost of the cheapest can so we're trying to minimize costs And therefore cost is going to be our optimizing function the statement about volume is the constraint here So the cost it looks like we're going to get five cents Times well the area of the top plus the area of the bottom And then we're going to you know cost 10 cents For the area of the side now given it's a cylinder the top and bottom of both circles So the area of the top and the bottom are actually the same So we're going to end up with 0.05 times two times the area of the top Plus 10 cents times the area of the side Which of course if you take five cents times by two that's 10 cents So we actually can figure out this 10 cents and we have to figure out the area of the top plus the area of the side Let's unravel this a little bit more. Well, like I already mentioned the top is a circle So the area of the circle is going to be pi r squared where r again is the radius of the circle here The side on the other hand Where what what is that shape going to be? Well, if you think of it as like the soup label on a can if you cut it off and flatten it You're going to get a rectangle for which one of the dimensions is going to be the height The other dimension will be the circumference of the circle here So we end up with two pi r and so area of a rectangle is length times width So you end up with two pi r h Like so So now we have that cost is a function of two variables r and h We need to remove one of the variables and that's where the constraint comes into play If you take your constraint and put divide both sides by pi We end up with r squared h is equal to 1000 And then if you divide both sides by r squared, you're going to get h equals 1000 Over r squared make that substitution into our cost function right here So cost is equal to 10 cents times pi r squared plus two pi r times 1000 Over r squared for which we can simplify this thing. I'm going to factor the pi out Of this situation here because we don't really need it for a while. You have an r squared You get two times a thousand which is 2000 r divided by r squared is going to be one over r But as I have to take the derivative soon, I'm going to write that as r to negative one power And this is our cost function to find the optimal cost. We need to take the derivative Now let's first think about the boundary for a second the the radius here could be allowed to go off towards zero That would make for a zero volume. So that doesn't quite work We could allow the height to go off towards zero. That's sort of the other extreme But that would also make volume equal to zero and it needs to be 1000 pi So these boundary values don't make any sense. So the optimal solution will not be on the boundary The optimal solution would have to be at a critical number for which we calculate that using just the usual derivative rules We're going to get 0.1 pi times 2 r plus I guess it's actually minus 2000 r to the negative 2 this needs to equal zero divide both sides by 0.1 pi We end up with 2 r minus 2000 r to the negative 2. I'm just going to write this as over r squared now equals zero We can move this friend to the other side. So we end up with 2 r equals 2000 Over r squared. I'm going to time both sides of the equation by r squared to get 2 r cubed equals 2000 Divide both sides by 2 we get r cubed equals 1000 and therefore take the cube root We get r equals the cube root of 1000 Which turns out to be 10 centimeters now what this tells you 10 centimeters. This is the optimal dimension for the for the arcane the radius should be 10 centimeters We could use this formula to figure out what the height should be if we wanted to we could plug it in 10 squared is 100 a thousand divided by Uh 100 is 10. So we want the radius to be 10. We want the height to be 10 That's going to be the minimal cost of arcane, but we're actually looking for the cost We don't have the cost right now. We need to plug this 10 centimeters into our cost function And start simplifying it from there, but be cautious With an optimization problem It's not always the case that the optimal solution is a critical number Sometimes it's the boundary, which is why we considered the boundary earlier Another way to see that this is the minimum cost is to do either the first or second derivative test Like if we calculate the second derivative of this function and plugged in x equals or excuse me r equals 10 We would end up with the second derivative being The second derivative being positive which therefore this would mean we have a minimum value or try to minimize cost here So let's plug this into our cost function the cost equals point 1 pi times 10 squared which is 100 plus 2000 over In this case r which is 10 which notice if you take a thousand two thousand divide by 10. That's just 200 200 plus 100 is 300 so you get point one pi times 300 And then point one times 300 will be 30. So we get an answer of 30 pi. What are the units here? Because I use sense Right to do 10 cents. I did point one. I'm actually working in dollars here And so my final answer would then be 30 pi dollars If you want to write it as sense, you'd at the times is by 100 and get 3000 Pi cents, but 30 dollars is typically the way we would do currency in the united states So that'll be our final answer if you want you could multiply that 30 by pi to get an answer and you round it to two decimal places That's very appropriate for again if you have a calculator for currency in the united states Uh, but assuming you don't have one you can leave it exact. You don't have a calculator that it is You can leave it exact and so the answer would be 30 pi So