Lagrange multiplier million dollar question. anybody use the method of lagrange multipliers to find the max and min values of g(s, t)=t^2e^s given that s^2+t^2=3 . how are we assured that these extreme values exist?

We are given the equation: g(s,t) = t^2 e^s

Which is subject to the constraint: s^2 + t^2=3

The points (x,y) that will maximize g(s,t) will be the points that will satisfy the equation ∇f(x, y, z) = λ∇g(x, y, z), therefore:

2 t e^s = λ (2 t), so  e^s = λ           --> 1

t^2 e^s = λ (2 s)                                                --> 2

s^2 + t^2 = 3                                       -->3

To solve this problem, note that λ cannot be zero by equation 1 since e^s can never be zero. Therefore plug in equation 1 to 2:

t^2 e^s = (e^s) (2 s)       

t^2 = 2 s                                               --> 4

Plug in equation 4 to 3:

s^2 + 2s = 3

By completing the square:

s^2 + 2s + 1 = 4

(s + 1)^2 = 4

s + 1 = ±2

s = -3, 1

Calculating for t using equation 4:

when s = -3

t^2 = 2(-3)

t = sqrt(-6)

Since t is imaginary, therefore s=-3 is not a solution

when s = 1

t^2 = 2(1)

t = sqrt(2) = ±1.414

Therefore the maxima and minima points are at:

(1, -1.414) and (1, 1.414)

g(1, -1.414)=(-1.414)^2 e^(1) = 5.434

g(1, -1.414)=( 1.414)^2 e^(1) = 5.434

x^6

step-by-step explanation:

this is correct because:

x-6=0

0*x=0

so

x^6