Okay, We're gonna be starting with a, um, right circular cylinder. So we have a right, um, circular cylinder, and it is topped by 1/2 of a, um, sphere. So it's taught by a hemisphere, Um, and this is actually a silo. So all silo is constructed by a I'm right. Circulars. Um, cylinder topped by a hemisphere. Um, so here is the radius of the hemisphere, which is also gonna be the radius of that circular cylinder. And we have the height of the cylinder and the cost of construction. Um, per square unit of the surface area is twice as great for the hemisphere, which makes sense because of the circular nature of that hemisphere. And we also know that the volume is fixed. So it's a fixed volume, and we want the cost of construction to be minimized. Okay. And so we do know that volume if we're gonna right the volume formula for these, the total volume is, um, the volume of that cylinder which is going to be pi r squared H plus half of a volume of a sphere which is 4/3 pi. Oh, are que and so this is gonna be pi r squared eight plus 2/3 pi r cute. Okay. And so we know costs of construction is going to be based off of the surface area. So let's come up with a surface area total surface area formula, so surface area, the total surface area is going to be and we only have to worry about the lateral area of that um, lado area of the cylinder. And so the lateral area of the cylinder is h times two pi are So that's gonna be the lateral area of the cylinder. Plus, Now we want half of the surface area of the him the sphere, because it's a hemisphere, and so that is only going to be too pie r squared. But we also know that it cost twice as much. So we need to multiply that by two. Okay. And so we want to minimize. So we know we're gonna have to take the derivative of the surface area. And so what we want to do is to right the surface area in terms of only one variable. And so we're gonna use this volume formula to solve for H because we know volume is fixed and So what we have here is that V minus 2/3 pi r cubed, divided by pi r squared is equal to H And so h h is gonna be equal to the over pi r squared minus kind of simplify this to third or and so over here in my surface area formula we're gonna now have two pi r times This volume over pi r squared mind is 2/3 are plus four pi r squared. And so now we're going to do some simplification. Try to get this in a better forms for us to take our derivative and so were you. Simplify this and we get this equal to To the over are minus four pi r squared over three plus four pi r squared And so let's go ahead and simplify a little bit more. And so let's go ahead and pull out, um r squared. And so we'll have an r squared here and where when we simplify that, that is going to be a pie. Over three kids will have a 12 pi minus a four pi Um and then this will be plus to the over our and so that's gonna be a whole lot easier to take the driven above. And so we take the derivative of the surface area with respect to our, which gives us to our times eight pi over three minus to V over r squared. And so let's go ahead and put that over a common denominator and we get 16 pi r cubed minus two V minus 60 so minus 60 because we're putting it over a common denominator of three r squared so 60 over three r squared. And so we know our cannot equal zero. And so now we want to set that derivative. We want to set that derivative equal to zero. Let's get back up there. Let's set that derivative equal to zero. So and so that's only gonna affect the numerator. So 16 pi r cubed minus six b. And so when we do that, we get our is equal to, um three B over eight. Hi to the 1/3. Okay, so that is what we think the radius is. And so let's see, um, if that will be a maximum or a minimum, um, place where my surface area will occur. So let's go ahead and take the um, second derivative. And so when we take thes second derivative, we get, um 16 16 pi over three plus four v over R cubed and we know are, of course has been greater than zero. So no matter what, this will be positive. Which means that this will be a place. Um, where surface area, um will be maximize or minimized. Sorry, cause it's positive. Which means it's Kong cave. Uh um And so now let's go ahead and find r h. So So we know that are is the three v the Cube ooh of three V over eight pi. We also know that h was also equal to, um V over see the minus the 2/3 pi times the, um, our cube. So this would be three V over eight pi divided by, um pie times are squared, and that's gonna be three V over eight pi to the 2/3 power. And so now we have this equal to the minus V over four, divided by that pie. Times three V over eight pi 2/3 power. So this is gonna be equal. Teoh three over four B and then we're gonna multiply both the top and bottom. We're gonna rationalize that denominator. So I have three V over eight pi to the 1/3 divided by the pie times the three V over eight pie. So H now becomes, um if I simplify three v over pie to the 1/3 and our is three V over eight pi to the 1/3. So they're my dimensions that will minimize the cost of the construction.