Find two numbers x and y such that a) 2x+y=100 and A=2x+2xy+y is maximized b) 2x+4y-15=0 and B= √x2+y2is minimized. Note that in this case, minimizing B is equivalent to minimizing B2.

We have to use Lagrange Multipliers to solve this problem. The maximum of a differentiable function f with the constraint g(x,y) = b, then we have that there exists a constant [tex] \lambda [/tex] such that

[tex] \nabla f(x,y) = \lambda \, \nabla g(x,y) [/tex]

Or, in other words,

[tex] f_x(x,y) = \lambda \, g_x(x,y) \\ f_y(x,y) = \lambda \, g_y(x,y) [/tex]

a) Lets compute the partial derivates of f(x,y) = 2x+2xy+y. Recall that, for example, the partial derivate of f respect to the variable x is obtained from derivating f thinking the variable y as a constant.

[tex] f_x(x,y) = 2 + 2y [/tex]

On the other hand,

[tex] f_y(x,y) = 2x+1 [/tex]

The restriction is g(x,y) = 100, with g(x,y) = 2x+y. The partial derivates of g are

[tex] g_x(x,y) = 2; g_y(x,y) = 1 [/tex]

This means that the Lagrange equations are

[tex]2y + 2 = 2 \, \lambda[/tex]
[tex]2x +1 = \lambda[/tex]
[tex]2x + y = 100[/tex] (this is the restriction, in other words, g(x,y) = 100)

Note that 2y + 2, which is [tex] 2 \, \lambda [/tex] is the double of 2x+1, which is [tex] \lambda [/tex]. Therefore, we can forget [tex] \lambda [/tex] for now and focus on x and y with this relation:

2y+2 = 2 (2x+1) = 4x+2

2y = 4x

y = 2x

If y is equal to 2x, then

g(x,y) = 2x+y = 2x+2x = 4x

Since g(x,y) = 100, we have that

4x = 100

x = 100/4 = 25

And, therefore y = 25*2 = 50

Therefore, x = 25, Y = 50.

b) We will use the suggestion and find the minumum of f(x,y) = B² = x²+y², under the constraing g(x,y) = 0, with g(x,y) = 2x+4y-15. The suggestion is based on the fact that B is positive fon any x and y; and if 2 numbers a, b are positive, and a < b, then a² < b². In other words, if (x,y) is the minimum of B, then (x,y) is also the minimum of B² = f.

Lets apply Lagrange multipliers again. First, we need to compute the partial derivates of f:

[tex]f_x(x,y) = 2x \\f_y(x,y) = 2y[/tex]

And now, the partial derivates of g:

[tex] g_x(x,y) = 2 \\ g_y(x,y) = 4 [/tex]

This gives us the following equations:

[tex] 2x = 2 \, \lambda \\ 2y = 4 \, \lambda \\ 2x+4y-15 = 0 [/tex]

If we compare 2x with 2y, we will find that 2y is the double of 2x, because 2y is equal to [tex] 4 \, \lambda [/tex] , while on the other hand, [tex] 2x = 2 \, \lambda [/tex] . As a consequence, we have

2y = 2*2x

y = 2x

Now, we replace y with 2x in the equation of g:

0 = g(x,y) = 2x+4y-15 = 2x+4*2x -1x = 10x-15

10 x = 15

x = 15/10 = 1.5

y = 1x5*2 = 3

Then, B is minimized for x 0 1.5, y = 3.