Find the maximum and minimum values of the function
Subject to the constraint .

D)
E)

Question

Find the maximum and minimum values of the function ​  ​
Subject to the constraint  .
​
D) ​ 
E) ​

Quizplus · Accepted Answer

We can use the method of Lagrange Multipliers to solve the problem. Let
f(x,y) = xy
g(x,y) = x² + y² - 1
We need to find the maximum and minimum values of f(x,y) subject to the constraint g(x,y) = 0.
The Lagrangian is:
L(x,y,λ) = f(x,y) - λg(x,y) = xy - λ(x² + y² - 1)
Taking partial derivatives and setting them equal to zero, we get:
∂L/∂x = y - 2λx = 0
∂L/∂y = x - 2λy = 0
∂L/∂λ = x² + y² - 1 = 0
Solving these equations, we get:
x = ±√(1/(2λ))
y = ±√(1/(2λ))
Substituting these values in g(x,y) = 0, we get:
1/(2λ) + 1/(2λ) = 1
λ = 1/√2
Using this value of λ, we get:
x = ±1/√2
y = ±1/√2
The function f(x,y) = xy takes the maximum value of 1/2 at (x,y) = (1/√2,1/√2) and the minimum value of -1/2 at (x,y) = (-1/√2,-1/√2).
Therefore, the answer is choice C: 11eb651f_cae3_2f62_b0a1_23f1d0074ad2_TB8255_11.

Find the Maximum and Minimum Values of the Function ​

Find the Maximum and Minimum Values of the Function