Chapter 14.8, Problem 1PT

To determine

The appropriate option for the point at which absolute maximum of $f\left(x,y\right)=12x-8y$ subject to $x^2+2y^2=11$ occurs from the given options $\left(\text{a}\right)\text{-}\left(\text{d}\right)$.

Answer to Problem 1PT

The appropriate option is $\underset{\_}{\left(\text{a}\right)}$.

Explanation of Solution

Definition used:

If $f\left(x,y\right)$ and $g\left(x,y\right)$ have continuous partial derivatives and $\left(a,b\right)$ is a local extremum for f when restricted to $g\left(x,y\right)=k$ (a constraint), then there is a number $\lambda$, called a Lagrange multiplier, such that: $\nabla f\left(a,b\right)=\lambda \nabla g\left(a,b\right)$.

Thus, solving the preceding constrained extremum problem is the same as solving the equations $\nabla f=\lambda \nabla g$ and $g\left(x,y\right)=k$.

Calculation:

Given function is $f\left(x,y\right)=12x-8y$.

Then, $\nabla f=\langle 12,-8\rangle$.

Let $g\left(x,y\right)=x^2+2y^2$.

Then, $\nabla g=\langle 2x,4y\rangle$.

Substitute $\nabla f=\langle 12,-8\rangle$ and $\nabla g=\langle 2x,4y\rangle$ in $\nabla f=\lambda \nabla g$.

$\begin{array}{l}\langle 12,-8\rangle =\lambda \langle 2x,4y\rangle \\ \langle 12,-8\rangle =\langle 2\lambda x,4\lambda y\rangle \end{array}$

Therefore, $12=2\lambda x$ and $-8=4\lambda y$.

Simplify $12=2\lambda x$ as follows.

$\begin{array}{l}12=2\lambda x\\ \lambda x=6\\ {\lambda }^2{x}^2=36\mathrm{......}\left(1\right)\end{array}$

Simplify $-8=4\lambda y$ as follows.

$\begin{array}{l}-8=4\lambda y\\ \lambda y=-2\\ {\lambda }^2{y}^2=4\left(\text{Multiply both sides by 2}\right)\\ 2{\lambda }^2{y}^2=8\mathrm{......}\left(2\right)\end{array}$

Add the equations (1) and (2).

$\begin{array}{l}{\lambda }^2{x}^2+2{\lambda }^2{y}^2=8+36\\ {\lambda }^2\left(x^2+2y^2\right)=44\\ {\lambda }^2\left(11\right)=44\\ {\lambda }^2=4\\ \lambda =2\end{array}$

Substitute $\lambda =2$ in $12=2\lambda x$.

$\begin{array}{l}12=2\left(2\right)x\\ 12=4x\\ x=3\end{array}$

Substitute $\lambda =2$ in $-8=4\lambda y$.

$\begin{array}{l}-8=4\left(2\right)y\\ y=\frac{-8}{8}\\ y=-1\end{array}$

Thus The absolute maximum of $f\left(x,y\right)=12x-8y$ subject to $x^2+2y^2=11$ occurs at $\left(3,-1\right)$.

Therefore, the appropriate option is $\underset{\_}{\left(\text{a}\right)}$.