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Numerade Educator



Problem 43 Hard Difficulty

A cone with the height $ h $ is inscribed in a larger cone with height $ H $ so that its vertex is at the center of the base of the larger cone. Show that the inner cone has a maximum volume when $ h = \dfrac{1}{3}H $.


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Video Transcript

we're told that a cone with height little H is inscribed in a larger cone with big H so that its vertex is at the center of the base of the larger cone. Whereas to show that the inner cone has a maximum volume and little H is one third of Big H Mhm, let's do a side view of the of the cones. Mhm this all? Yeah, just like artist First, I did find some variables. I'll let big R B radius of the larger cone and Big H is the height of larger cone were already told this while little R is the radius of the little cone and little H is the height of the little cone, right? Yeah, What now I'll draw the larger cone with a smaller cone inside it. Gucci and Lucci? Yeah, yeah, it's bushy and popping bows in Tokyo, alleging seven elmerson. It's like right first we have a It's right triangle here. Now, looking at this figure, we actually see similar triangles, so we have the larger height h over the larger radius R. This is the same as the smaller height middle age over the smaller radius. Little are I'm sorry. Not quite. It's the difference between heights. Big H minus little h over a little r because we have this larger triangle here racist and the smaller triangle inside of it references. Okay, so we'll solve for little h, we get that little H equals big H over our times. Bigger r minus. Little are. Mm. Yeah. Now let's let's do this result into the equation for volume. So the volume of the smaller cone? Yeah, this is one thirds pi r squared little h and so we get one third pie r squared times big H over big are times bigger r minus little are And so we can express the volume as a function of our in this way. Volume of little are this is one third, 308. Yeah. Pi r squared each. Mhm, yeah. Minus pi r squared h little are over three Big are okay. This is the same as pi r cubed big H over three. Big are now differentiate respect to our Yeah, The prime of art is two pi r gig h over three minus pi r squared. Big h uber big are we want to find when this is equal to zero. It's a K. So I can factor this as pi r h over three. Big are okay, and I have left over. This is too big. Are minus Okay, three little are equals. Zero, this adds. And so it follows that either R equals zero or R is equal to two. Big are over 23 The radius of the smaller cone can't be zero. So really only considering our second critical value. Now, notice that the factor two ar minus three little are is in fact, greater than zero If and only if little are is less than two. Big are over three. And to our minus three little R is less than zero if and only if little R is greater than two are over three therefore, yeah, by the first derivative test, it follows that our volume the has a maximum at r equals mm two big are over three. And so the height of the smaller cone at this level for our little age. Looking at a previous equation, this is a big H over big R times R minus little R, which is too big are over three. In fact, throughout the big are we get h times one minus two thirds, which is big H over three. Yes,