# Math Help - calc 3, maxes and mins. quick question

1. ## calc 3, maxes and mins. quick question

find the maximum and minimum values of the function subject to the given contraint.
f(x,y)= 8x+3y
subject to (x-1)^2 + (y+2)^2 = 9

so i took the gradient of each..
(8, 3)
(2x-2, 2y+2)

and to solve for the max and min, we need to use lagrange multipliers. so the gradient of f is equal to the gradient of g times lambda (some constant)

i have the equations
8= lambda(2x-2)
3=lambda(2y+4)
(x-1)^2 + (y+2)^2 =9

how do i solve this system of equations? the answer in the back of the book is kind of crazy...
x= 1 - 24/sqrt 73
y= -2- 9/sqrt 73
lambda= -sqrt 73/6

2. Originally Posted by holly123
find the maximum and minimum values of the function subject to the given contraint.
f(x,y)= 8x+3y
subject to (x-1)^2 + (y+2)^2 = 9

so i took the gradient of each..
(8, 3)
(2x-2, 2y+2)

and to solve for the max and min, we need to use lagrange multipliers. so the gradient of f is equal to the gradient of g times lambda (some constant)

i have the equations
8= lambda(2x-2)
3=lambda(2y+4)
(x-1)^2 + (y+2)^2 =9

how do i solve this system of equations? the answer in the back of the book is kind of crazy...
x= 1 - 24/sqrt 73
y= -2- 9/sqrt 73
lambda= -sqrt 73/6
From the first two

$x - 1 = \frac{8}{2 \lambda},\;\; y + 2 = \frac{3}{2 \lambda}$

then subs. into the third $(x-1)^2 +(y+2)^2 = 9$. This give you an equation for $\lambda$ which you can solve.

3. i'm working on the next problem and having the same problem as above. i have the equations:
-4x= lambda x
-18y= 2lambda y
8= lambda z
(x^2)/4 + (y^2)/9 + (z^2)/16=1

how do i solve for x and y?

4. Originally Posted by holly123
i'm working on the next problem and having the same problem as above. i have the equations:
-4x= lambda x
-18y= 2lambda y
8= lambda z
(x^2)/4 + (y^2)/9 + (z^2)/16=1

how do i solve for x and y?
From the first you have $(\lambda+4) x = 0$ and from the second $2(\lambda + 9) y = 0$ which gives rise to three cases:

$
(i) \; x = 0, \;y= 0
$

$
(ii) \; x = 0,\; \lambda = -9,
$

$
(iii)\; y = 0,\; \lambda = -4.
$

Look at each separately.

5. so the first one would be x=o or lambda= -4
then lambda=-9 or y=0
lambda= 8/z

is that right? then what do i do? sorry this part is just not clicking for me...

6. okay so i tried plugging into the last equation
so 0 + 0 + (8/lambda)^2/16 = 1
so lambda= +- 4
is this correct or totally wrong?