Question

Consider the two functions \( f(x,y) = x + y \) and \( g(x,y) = xy - 16 \) defined on \( \mathbb{R}^2 \). Then

• A. The function \(f\) has no global extreme value subject to the condition \( g = 0. \)
• B. The function \(f\) attains global extreme values at \((4,4)\) and \((-4,-4)\) subject to the condition \( g = 0. \)
• C. The function \(g\) has no global extreme value subject to the condition \( f = 0. \)
• D. The function \(g\) has a global extreme value at \((0,0)\) subject to the condition \( f = 0. \)

Hint:

Lagrange multipliers are used for constrained optimization. Here, the symmetry of \(f\) and \(g\) simplifies the condition to \(x=y.\)

Answer 1

The Correct Option is A, D

Step 1: Constraint equation.
The constraint \(g(x,y)=0\) gives \(xy=16.\) We need to find the extrema of \(f(x,y)=x+y\) subject to this condition.
Step 2: Apply Lagrange multipliers.
Let \( \nabla f = \lambda \nabla g.\) Then \[ (1, 1) = \lambda (y, x). \] This gives \(1 = \lambda y\) and \(1 = \lambda x \Rightarrow x = y.\)
Step 3: Use the constraint.
If \(x = y,\) then \(x^2 = 16 \Rightarrow x = \pm 4.\) Hence, points \((4,4)\) and \((-4,-4)\) satisfy the condition.
Step 4: Check extrema.
At \((4,4)\), \(f = 8\); at \((-4,-4)\), \(f = -8.\) Thus, both are global extrema.
Step 5: Conclusion.
(B) is correct.