# Use Lagrange multipliers to find the maximum(s) and minimum(s)

Use Lagrange multipliers to find the maximum(s) and minimum(s) of

$$f(x,y)=2x^2+3y^2−4x−5 \text{ subject to } x^2+y^2=16$$

So far I've taken the partials of f and g:

\begin{align} f_x &= 4x-4 \\ f_y &= 6y \\ g_x &= 2x \\ g_y &= 2y \end{align}

Then I set them equal to each other and multiplied by $\lambda$: \begin{align} 4x-4 &= 2x\lambda \\ 6y &= 2y\lambda \end{align}

I'm kind of stuck here where I need to solve for $\lambda$.

• just use partial derivatives Mar 7 '14 at 0:50
• If $6y = 2y\lambda$, what's $\lambda$?
– dfan
Mar 7 '14 at 1:08
• 3, so would that mean x = -2? Also, if that is correct, do you know how I could solve for y? thanks Mar 7 '14 at 1:10
• Don't forget that you have three constraints to find your three variables $x$, $y$, and $\lambda$: the two equations you got from equating partial derivatives, and the original constraint $x^2+y^2=16$.
– dfan
Mar 7 '14 at 1:12
• In Lagrange multiplier problems it is vital to find all solutions of your equations. So $6y=2y\lambda$ does not tell you that $\lambda=3$, it tells you that either $\lambda=3$ or $y=0$. Mar 7 '14 at 1:13

Solve for $y^2: y^2 = 16 - x^2 \to f(x,y) = 2x^2 + 3(16 - x^2) - 4x - 5 = -x^2 - 4x + 43 = -(x + 2)^2 + 47 = g(x)$

Consider $f(x,y) = g(x) = 47 - (x + 2)^2$ on $[-4, 4]$. Clearly:$g(x) \leq 47$ and that $\max g(x) = 47$ when $x = -2$ so $y^2 = 16 - (-2)^2 = 12$, and $y = \sqrt{12}$ and$y = - \sqrt{12}$. Observe that $g$ is minimized when $(x + 2)^2$ is maximized, and this occurs when $x = 4$. So $\min g = 47 - (4 + 2)^ = 47 - 36 = 11$.

• Thank you for the help! Makes sense now. Mar 7 '14 at 1:58

In this problem, the constraint curve $$\ x^2 \ + \ y^2 \ = \ 16 \$$ has four-fold symmetry about the origin, but the function $$\ f(x,y) \ = \ 2x^2 \ + \ 3y^2 \ − \ 4x \ − \ 5 \$$ is only symmetrical about the $$x-$$ axis. This indicates that we can expect critical points on the constraint circle to appear in pairs $$\ (x,y) \$$ and $$\ ( x, -y) \$$ .

For your Lagrange equations, the one involving $$\ x \$$ does not have a convenient factorization after being rearranged, but the one with $$\ y \$$ does:

$$4x \ - \ 4 \ = \ \lambda·2x \ \ \rightarrow \ \ (2 \ - \ \lambda)·x \ = \ 2 \ \ \ , \ \ \ 6y \ = \ \lambda·2y \ \ \rightarrow \ \ (3 \ - \ \lambda)·y \ = \ 0 \ \ .$$

$$\mathbf{y = 0 : } \quad x^2 \ = \ 16 \ \ \Rightarrow \ \ x \ = \ \pm 4 \ \ ,$$ giving us the two function values

$$\ f(4,0) \ = \ 2·4^2 \ − \ 4·4 \ − \ 5 \ \ = \ \ 11 \ \ \ \text{or} \ \ \ f(-4,0) \ = \ 2·(-4)^2 \ − \ 4·(-4) \ − \ 5 \ \ = \ \ 43 \ \ ;$$

or, applying your Lagrange equation for $$\ x \$$ ,

$$\mathbf{\lambda = 3 : } \quad (2 \ - \ 3)·x \ = \ 2 \ \ \Rightarrow \ \ x \ = \ -2 \ \ \Rightarrow \ \ (-2)^2 \ + \ y^2 \ = \ 16 \ \ \Rightarrow \ \ y^2 \ = \ 12 \ \ \Rightarrow \ \ y \ = \ \pm 2\sqrt{3}$$ $$f(-2 \ , \ \pm 2\sqrt{3}) \ = \ 2·(-2)^2 \ + \ 3·12 \ − \ 4·(-2) \ − \ 5 \ \ = \ \ 47 \ \ ,$$

this latter result illustrating the function symmetry mentioned above.

The main aim here in summarizing the earlier work is to show the geometric interpretation of the problem, which also allows two of the results to be found with relatively little calculation. If we consider level curves of our function $$\ 2x^2 \ + \ 3y^2 \ − \ 4x \ − \ 5 \ \ = \ \ c \ \ ,$$ we can re-write this equation as

$$2x^2 \ − \ 4x \ + \ 3y^2 \ \ = \ \ 5 \ + \ c \ \ \rightarrow \ \ 2·(x^2 \ − \ 2x \ + \ 1 ) \ + \ 3y^2 \ \ = \ \ 5 \ + \ 2 \ + \ c$$ $$\rightarrow \ \ \frac{(x \ − \ 1 )^2}{\left( \frac{7+c}{2} \right)} \ + \ \frac{y^2}{\left( \frac{7+c}{3} \right)} \ \ = \ \ 1 \ \ .$$

These level curves are ellipses centered on $$\ (1,0) \$$ with their major axes on the $$\ x-$$ axis; the graph below shows the ellipse for $$\ c \ = \ 0 \$$ and the constraint circle.

Critical values of $$\ f(x,y) \ = \ c \$$ occur when the endpoints of the major axis are tangent to the circle, that is, when either the right endpoint lies at $$\ (4,0) \$$ or the left endpoint becomes $$\ (-4,0) \$$ . In the first case, the distance from the center of the ellipse to the right-hand intersection of the circle with the $$\ x-$$ axis is $$\ 3 \$$ , so the semi-major axis of the ellipse must be $$\ \sqrt{\frac{7+c}{2}} \ = \ 3 \ \Rightarrow \ 7 + c \ = \ 2·9 \ \Rightarrow \ c \ = \ 11 \ .$$ The left $$\ x-$$ intercept of the circle is $$\ 5 \$$ units from the ellipse's center, so we obtain $$\ \sqrt{\frac{7+c}{2}} \ = \ 5 \ \Rightarrow \ 7 + c \ = \ 2·25$$ $$\Rightarrow \ c \ = \ 43 \ .$$ The corresponding graphs are displayed below.

This does not seem quite complete, however, as it appears that if the ellipse were made slightly larger than the $$\ c \ = \ 43 \$$ level curve, we would have the circle circumscribed by the ellipse.

EDIT: We can locate the remaining tangent points by inserting the constraint condition into the level curve equation:

$$2x^2 \ + \ 3·(16-x^2) \ - \ 4x \ - \ 5 \ \ = \ \ c \ \ \Rightarrow \ \ -x^2 \ - \ 4x \ \ = \ \ c \ - \ 43 \ \$$ $$\Rightarrow \ \ (x \ + \ 2)^2 \ \ = \ \ 47 \ - \ c \ \ \Rightarrow \ \ x \ \ = \ \ -2 \ \pm \ \sqrt{47-c} \ \ .$$

What would be four intersection points of the level curve and constrain circle "coalesce" into two tangent points for $$\ c \ = \ 47 \ .$$ We thus have the result $$\ x \ = \ -2 \$$ for the tangent points that we obtained from the $$\ \lambda \ = \ 3 \$$ case.