Find the maximum and minimum values of the function subject to the constraint .
A) Maxima : Minima :
B) Maxima : Minima :
C) Maxima : Minima :
D) Maxima : Minima :

Question

Find the maximum and minimum values of the function  subject to the constraint  .
A) Maxima : Minima : 
B) Maxima : Minima : 
C) Maxima : Minima : 
D) Maxima : Minima :

Quizplus · Accepted Answer

The function is subject to the constraint 11eaa8b1_d993_2c1e_b1b6_9b6f7e18b3ff_TB6026_11, which is equivalent to a circle centred at (1,1) with radius 1. 
Using Lagrange multipliers, we can find the maxima and minima of the function subject to this constraint.
We have:
∇f = λ∇g
⟺ (4x-4) = λ(2x-2)
⟺ 2x-2 = λx - λ
Similarly, we have:
2y-2 = λy - λ
Solving these equations, we get:
λ = 2x-2x+2 = 4
λ = 2y-2y+2 = 2
Thus,
x = 3/2 and y = 3/2
Checking the Hessian, we see that we have a maximum at this point.
Plugging in these values into the function, we get:
f(3/2,3/2) = 11eaa8b1_d993_7a3f_b1b6_f1238ac4ea51_TB6026_11
As for the minimum, we note that the constraint is a closed and bounded set, so by the extreme value theorem, the function attains a minimum somewhere on this circle. Since the function is symmetric around the line x=y, the minimum must occur at (1,1), where the function takes on the value:
f(1,1) = 11eaa8b1_d993_7a40_b1b6_8132b789d7c7_TB6026_11

Find the Maximum and Minimum Values of the Function