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