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 831: 12

Answer

There is a critical point at $\left( {0, - \frac{1}{2},\frac{1}{4}} \right)$. The critical value is $f\left( {0, - \frac{1}{2},\frac{1}{4}} \right) = \frac{1}{4}$. However, this value does not correspond to the maximum nor minimum values of $f$ along the constraint.

Work Step by Step

We have $f\left( {x,y,z} \right) = {x^2} - y - z$ and the constraint $g\left( {x,y,z} \right) = {x^2} - {y^2} + z = 0$. Step 1. Write out the Lagrange equations Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields $\left( {2x, - 1, - 1} \right) = \lambda \left( {2x, - 2y,1} \right)$ $2x = 2\lambda x$, ${\ \ }$ $ - 1 = - 2\lambda y$, ${\ \ }$ $ - 1 = \lambda $ Step 2. Solve for $\lambda$ in terms of $x$ and $y$ From Step 1, we obtain $\lambda = - 1$. Step 3. Solve for $x$ and $y$ using the constraint Substituting $\lambda = - 1$ in the equations in Step 1, we obtain $x=0$, $y = - \frac{1}{2}$. Substituting $x=0$ and $y = - \frac{1}{2}$ in the constraint $g\left( {x,y,z} \right)$ gives $ - {\left( { - \frac{1}{2}} \right)^2} + z = 0$ $z = \frac{1}{4}$ So, the critical point (the red point in the figure attached) is $\left( {0, - \frac{1}{2},\frac{1}{4}} \right)$. Step 4. Calculate the critical values The value of $f$ at the critical point is $f\left( {0, - \frac{1}{2},\frac{1}{4}} \right) = \frac{1}{4}$. Referring to the figure attached, we see that the level surface of $f$ whose value is $\frac{1}{4}$, intersects the constraint $g\left( {x,y,z} \right) = 0$ at the critical point $\left( {0, - \frac{1}{2},\frac{1}{4}} \right)$. However, we also notice that there are level surfaces of $f$ that have smaller values than $\frac{1}{4}$ and there are level surfaces of $f$ that have larger values than $\frac{1}{4}$ intersect the constraint $g\left( {x,y,z} \right) = 0$. Thus, the level surface whose value is $\frac{1}{4}$ is neither maximum nor minimum of $f$ along the constraint $g\left( {x,y,z} \right) = 0$. So, we conclude that $f\left( {0, - \frac{1}{2},\frac{1}{4}} \right) = \frac{1}{4}$ does not correspond to the maximum nor minimum values of $f$ along the constraint.
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