Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 15 - Differentiation in Several Variables - 15.8 Lagrange Multipliers: Optimizing with a Constraint - Exercises - Page 832: 33

Answer

$\left( {r,h} \right) = \left( {3,6} \right)$

Work Step by Step

We have the surface area (including the top and bottom): $S\left( {r,h} \right) = 2\pi rh + 2\pi {r^2}$ Our task is to minimize $S$ subject to the constraint $V = \pi {r^2}h = 54\pi $. Step 1. Write out the Lagrange equations Using Theorem 1, the Lagrange condition $\nabla S = \lambda \nabla V$ yields $\left( {2\pi h + 4\pi r,2\pi r} \right) = \lambda \left( {2\pi rh,\pi {r^2}} \right)$ So, the Lagrange equations are $2\pi h + 4\pi r = \lambda \left( {2\pi rh} \right)$, ${\ \ \ }$ $2\pi r = \lambda \left( {\pi {r^2}} \right)$ (1) ${\ \ \ \ }$ $h + 2r = rh\lambda $, ${\ \ \ }$ $2 = r\lambda $ Step 2. Solve for $\lambda$ in terms of $r$ and $h$ The second equation of (1) implies that $r \ne 0$ and $\lambda \ne 0$. As a result $h \ne 0$. From equation (1) we obtain $\lambda = \frac{2}{r} = \frac{{h + 2r}}{{rh}}$ $2 = 1 + 2\frac{r}{h}$ $1 = 2\frac{r}{h}$ So, $h = 2r$. Step 3. Solve for $r$ and $h$ using the constraint Substituting $h = 2r$ in the constraint gives $\pi {r^2}\left( {2r} \right) = 54\pi $ ${r^3} = 27$ $r = 3$ The critical point is $\left( {r,h} \right) = \left( {3,6} \right)$. Step 4. Calculate the critical values Using the critical point $\left( {r,h} \right) = \left( {3,6} \right)$ we evaluate the extreme value of $S$ subject to the constraint $V = \pi {r^2}h = 54\pi $: $S\left( {3,6} \right) = 54\pi $ For a given constraint $V = 54\pi $ we may increase $r$ and at the same time decrease $h$ such that $V$ is constant. Since $S$ is increasing, we conclude that $S\left( {3,6} \right) = 54\pi $ is a minimum value subject to the constraint.
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