This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint. f(x, y) = x2 − y2; x2 + y2 = 4

Step-by-step explanation:

f(x,y) = x²-y² and the restriction is

g(x,y) = 4

with g(x,y) = x²+y²

an extreme (x0,y0) of f following that restriction must satisfy

For ceratin constant λ

Now

Thus, they are equal in the first coordinate, so in order for them to be proportional one of the following must happen:

So, the critical points have the form (x0,0) and (0,y0), also since they have to fulfill the restriction, we have

x0² + 0² = 4, then x0 = 2 or -2

same with (0,y0)

0²+y0² = 4, then y0 must be 2 or -2

Then the critical points are (2,0), (-2,0), (0,2), (0,-2).

In order to find the extremes now we just have to evaluate in each candidate we have and compare the results

f(2,0) = 2² = 4

f(-2,0) = (-2)² = 4

f(0,2) = 0-2² = -4

f(0,-2) = 0-(-2)² = -4

Thus

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions