Question

Use Lagrange multipliers to find the absolute maximum value and absolute minimum value of f(x, y) = 2x 2 + 2y subject to x 2 + 2y + y 2 = 15.

Answer 1

Answer:

Maximum absolute value of f=61/2 at ($3/2\sqrt{7},-1/2$) and ($-3/2\sqrt{7},-1/2$)

Minimum absolute value of f=-10 at (0,-5)

Step-by-step explanation:

We are given that

$f(x,y)=2x^2+2y$

Let

$g(x,y)=x^2+2y+y^2-15=0$

$\nabla f(x,y)=<4x,2>$

$\nabla g(x,y)=<2x,2+2y>$

Using Lagrange multipliers

$f_x(x,y)=\lambda g_x(x,y)$

$4x=2x\lambda$

$4x-2x\lambda=2x(2-\lambda)=0$

x=0 or

$\lambda=2$

If x=0

Then,

$y^2+2y-15=0$

$y^2+5y-3y-15=0$

$y(y+5)-3(y+5)=0$

$(y+5)(y-3)=0$

$y=-5,3$

$2=(2+2y)\lambda$

If $\lambda=2$

$2=2(2+2y)$

$2/2=2+2y$

$1=2+2y$

$2y=1-2=-1$

$y=-\frac{1}{2}$

If y=-1/2

Then,

$x^2-1+1/4=15$

$x^2+(\frac{-4+1}{4})=15$

$x^2-\frac{3}{4}=15$

$x^2=15+3/4=\frac{60+3}{4}=\frac{63}{4}$

$x=\pm\frac{3\sqrt{7}}{2}$

Therefore, possible extreme points are

(0,-5),(0,3),($3/2\sqrt{7},-1/2$) and ($-3/2\sqrt{7},-1/2$)

$f(0,-5)=2(0)+2(-5)=-10$

$f(0,3)=2(3)=6$

$f(3/2\sqrt{7},-1/2)=\frac{63}{2}-1=\frac{61}{2}$

$f(-3/2\sqrt{7},-1/2)=\frac{63}{2}-1=\frac{61}{2}$

Therefore, maximum absolute value of f=61/2 at ($3/2\sqrt{7},-1/2$) and ($-3/2\sqrt{7},-1/2$)

Minimum absolute value of f=-10 at (0,-5)