Lagrange Multiplier Find Maximum and Minimum

Question

Use Lagrange multiplier to find maximum and minimum of $f(x,y) = 3x-4y$ subject to $x^2+3y^2=129$.

So I start by getting the partial with respect to both f(x, y) and $x^2+3y^2=129$

Ultimately, I get $f x(x, y) = 3$ $f y(x, y) = -4$ $g x(x, y) = 2x$ $g y(x, y) = 6y$

Then I equate them to get $3 = 2x(times constant)$ and $-4 = 6y(times constant)$

How do I simplify to get x and y. I tried solving for both the constant and the variables but wasn't getting a correct answer. Can anyone assist?

Answer 1

$$ [3,-4]=\lambda[2x,,6y]=[2x\lambda,6y\lambda] $$ $$ 3=2x\lambda, -4= 6y\lambda $$ $$ \lambda =3/(2x) = -4/(6y)= -2/(3y) $$ $$ 9y=-4x $$ Now substitute $ y=(-4/9)x $ into the constraint.

Answer 2

What you have described is not exactly the Lagrange Multiplier method. Consider this guide for a start.

Lagrange Multiplier method

Step 1

Identify your function and your constraint equations. There may be more that one constraint equation but the function is always one. Of course, the function may be given by several equations or in piecewise form.

In this case, the function is $f(x,y) = 3x - 4y$, and there is one constraint equation $x^2 +3y^2 = 129$

Step 2

Set up the constraint equations such that they have a zero on the right-hand side. In this case, $x^2 +3y^2 = 129$ becomes $x^2 +3y^2 - 129 = 0$. This is optional.

Next, multiply the constraint equations by unique constants. The convention is to use $\lambda$'s. If there are several constraints, for instance, multiply the first by $\lambda_1$, the second by $\lambda_2$ and so on. Just make sure the $\lambda$'s are different for each constraint.

Step 3

Create the Lagrange equation for the given system by setting equating the given function to the sum of the equations from step 2. The general form, for two-variable functions, (where $c_i(x,y)$ means the $i$th constrant equation) will look like this:

$$L(x,y, \lambda_1, \lambda_2, \ldots ) : f(x,y)= \lambda_1(c_1(x,y)) + \lambda_2(c_2(x,y)) + \lambda_3(c_2(x,y)) + \ldots $$

You can change the sign of the $\lambda$'s, as long as you don't alternate.

Back to your case, the Lagrange equation turns out to be

$$L(x,y,\lambda_1) : 3x - 4y = \lambda_1(x^2 + 3y^2-129) \\ \implies L(x,y,\lambda_1) = 3x - 4y - \lambda_1(x^2 + 3y^2-129)$$

Step 4

Find the partials of $L$ with respect to the initial variables and equate them to zero. Then solve the system of equations, taking into account the constraint equations. This will give you the critical points. Note: since there is only one constraint in your case, we'll just use $\lambda$ to represent $\lambda_1$.

$${L^{\prime}}_x = 0 \implies 3 - 2\lambda x = 0 \\ {L^{\prime}}_y = 0 \implies -4 - 6\lambda y = 0 $$

So you have the following system:

$$\begin{cases} 3 - 2\lambda x = 0 \\ -4 - 6\lambda y = 0 \\ x^2 +3y^2 - 129 = 0 \end{cases}$$

From the first and second equations $$3 - 2\lambda x = 0 \implies \lambda = \dfrac{3}{2x} \\ -4 - 6\lambda y = 0 \implies \lambda = \dfrac{-2}{3y} \\ \therefore \dfrac{3}{2x} = \dfrac{-2}{3y} \implies x = -\frac{9y}4$$

Then plug that into the constraint equation:

$$\left(-\frac{9y}4\right)^2+ 3y^2 - 129 = 0 \implies y = \pm 4 $$

That gives you two points $(-9, 4)$ and $(9,4)$.

What is left is to verify which of them is a maximum, a minimum or neither. Can you proceed?