Optimal soda can

a. Classical problem Find th...

Question

Optimal soda can

a. Classical problem Find the radius and height of a cylindrical soda can with a volume of 354 cm³ that minimize the surface area.

Accepted Answer

Welcome back, everyone. In this problem, an engineer is designing a cylindrical water tank to hold 5000 cubic centimeters of water, determine the radius of the tank that will minimize the surface area. A says it's 5 multiplied by a route 320 divided by pi square centimeters. B 5 multiplied by route 320 centimeters. C 5 route 320 centimeters and D 5 route 3 of 20 divided by pi centimeters. Now how are we going to find a radius that will minimize the surface area? Well, let's recall what we know about surface area. For a cylinder, recall that its total surface area includes the lateral surface and two circular bases, OK? So this is what our cylinder would look like. So if our cylinder has a height of age. And a radius of R2, OK? Then it's total surface area A would be equal to 2 RH plus 2 Rsquared. Now if we're trying to minimize the surface area, it would be best to have the area in terms of one variable. But in this case, we're expressing A in terms of R and H. So can we express H possibly in terms of R? Well, since our volume is fixed at 5000 cubic centimeters, OK. And we know. That volume is equal to pi R2 H, OK. Then that means i squared would be equal to 5000 and thus H. Is going to be equal to, then, 5000 divided by pi R squared. Since H equals 5000 divided by pi are squared, no we can express. Or surface area as a function of the radius, which means it's gonna be equal to. Let me write that properly here, therefore. A of R is gonna be equal to 2 R. Multiplied by 5000 divided by pi are squared. Plus 2 pi R squared, OK? And now when we simplify that, it's going to be equal to 10,000 divided by R plus 2 pi R squared. Know that we have a purely in terms of R, we can differentiate A of R and set it to 0, OK? So let me actually, let me write that in red. Just to show that we're doing something different here. So now we're differentiating A of R. With respect to I. OK. And now when we differentiate it and set it to zero, we can solve for a critical point or a critical value of R and determine if that value is what will minimize the surface area. Now, when we differentiate AFR. Notice here we have two terms. The derivative of 10,000 divided by R is going to be negative 10,000 multiplied by R to the negative 2, and the derivative of 2 squared is going to be 4 R. So now that we have our derivative, which we could even rewrite as negative 10,000. Divided by R2 + 4 by R, OK. Then when our first derivative equals 0. Then we can set this to 0. OK, so we'll have negative 10,000, OK, divided by R squared. Plus 4 R to be equal to 0. Now, we could combine this, OK, as, sorry, 4R. We could combine this as one term. And now that's gonna be when we do that. Let me make some space here. This is gonna be negative 10,000 plus 4 pi R cubed divided by R2 equals 0. Of course, where R is not equal to 0 because that would make our function undefined. Now, since we've set this to zero, the derivative will be 0 when when the numerator is 0. So this implies that negative 10,000 plus 4 pi R cubed equals 0, OK? Which means then. Now we can say 40 RQ will be equal to 10,000. Thus, if we divide through by 4 pi, then our cubed will be equal to 2500 divided by pi, which means that R. OK, is going to be equal to. The cube root, the cube root. Of 2500 divided by pi centimeters. And now, using what we know about roots, OK, let me move over here. We can simplify this even further to say that R will be equal to 5 multiplied by root 3 of 20 divided by pi centimeters, which means R is approximately 9.27 centimeters. Now that we have a value for R, the question is, is this value a minimum or maximum? Well, we can confirm that by using the derivative test, the first derivative test. So let's confirm that our value for R. Corresponds to a minimum. Using the first derivative test. Now the first derivative test basically tells us that if the sign changes from negative to positive for values before and after our critical point, then it's a minimum, but if it changes from positive to negative, it's a maximum. So let's test those two scenarios. First, for our less than our value of 520 divided by pi centimeters. Let's test if, or let's try to figure out the sign here for R, OK? So, no, we could choose a value for R and in this case, let's let R be equal to 9 because we know that our critical point is approximately 9.27 centimeters. So if we let Are benign centimeters. Then our first derivative. Is going to be equal to -10000. Plus, whereby multiplied by 9 cubed. Which is gonna be approximately -839. At 8.39, negative 839.12. OK. No, For a value of are greater than 5 cub 20 divided by 5 centimeters. Let's let R be tend. And now when R is 10 centimeters, then our first derivative of the surface area is going to be negative 10,000 plus 4 pi multiplied by 10 cubed, which is approximately 2,566.37. Now what do we notice here? Well, since the derivative sign changes from positive to negative, that means our critical point R is 5 multiplied by the cube root of 2, let me write that properly here 5 multiplied by the cube root of 20 divided by pi centimeters. Is a minimum, and we're sure it's a minimum because the sign changes from negative to positive. Therefore, if we look back at our answer choices, that means our correct answer for R is D. Thanks a lot for watching everyone. I hope this video helped.

Optimal soda cana. Classical problem Find the radius and height of a cylindrical soda can with a volume of 354 cm³ that minimize the surface area.

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Optimal soda can

a. Classical problem Find the radius and height of a cylindrical soda can with a volume of 354 cm³ that minimize the surface area.