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Exercise 7.8 · Q11

Q.The volume of a cylinder is given by the formula V=pir2hV=\\pi r^2h. Find the greatest and least values of VV if r+h=6r+h=6.

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Use r+h=6r+h=6 to write VV in terms of rr alone on the closed interval [0,6][0,6], then apply the Extreme Value Theorem procedure (critical numbers and endpoints).

Step 1. Set up. h=6−rh=6-r, with 0≤r≤60\le r\le6 (both radius and height must be non-negative).

V(r)=πr2h=πr2(6−r)=π(6r2−r3).V(r)=\pi r^2h=\pi r^2(6-r)=\pi(6r^2-r^3).

Step 2. Differentiate and solve V′(r)=0V'(r)=0.

V′(r)=π(12r−3r2)=3πr(4−r)=0 ⇒ r=0 or r=4.V'(r)=\pi(12r-3r^2)=3\pi r(4-r)=0\ \Rightarrow\ r=0\text{ or }r=4.

Step 3. Evaluate VV at the critical number and both endpoints.

V(0)=0V(0)=0.  V(4)=π(16)(2)=32π\ V(4)=\pi(16)(2)=32\pi.  V(6)=π(36)(0)=0\ V(6)=\pi(36)(0)=0.

Step 4. Compare. …

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