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Exercise 7.8 · Q12

Q.A hollow cone with base radius aa cm and height bb cm is placed on a table. Show that the volume of the largest cylinder that can be hidden underneath is dfrac49\\dfrac49 times volume of the cone.

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Use similar triangles to relate the cylinder's radius and height under the cone, reduce the cylinder's volume to one variable, maximize it, and compare to the cone's own volume.

Step 1. Relate cylinder radius xx and height yy via similar cones.

The cone has base radius aa, height bb. A cylinder of radius xx inscribed with its top touching the cone's slanted side, at height yy from the table, satisfies (by similar triangles, since the cone's radius shrinks linearly from aa at the base to 00 at the apex):

b−yb=xa ⇒ y=b(1−xa).\frac{b-y}{b}=\frac{x}{a}\ \Rightarrow\ y=b\left(1-\frac{x}{a}\right).

Step 2. Write the cylinder's volume in terms of xx alone.

V(x)=πx2y=πx2⋅b(1−xa)=πb(x2−x3a).V(x)=\pi x^2y=\pi x^2\cdot b\left(1-\frac{x}{a}\right)=\pi b\left(x^2-\frac{x^3}{a}\right).

Step 3. Differentiate and solve V′(x)=0V'(x)=0.

V′(x)=πb(2x−3x2a)=πbx(2−3xa)=0 ⇒ x=0 or x=2a3.V'(x)=\pi b\left(2x-\frac{3x^2}{a}\right)=\pi bx\left(2-\frac{3x}{a}\right)=0\ \Rightarrow\ x=0\text{ or }x=\frac{2a}{3}.

The nonzero critical number is x=2a3x=\dfrac{2a}3 (the maximizing radius, confirmed by checking VV rises then falls, or via V′′<0V''<0 there).

Step 4. Find the corresponding height and maximum cylinder volume. …

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