Find the absolute maximum and minimum values of f on the set D. f(x, y)= xy², D = {(x, y)| x >= 0, y >= 0, x² + y² <= 3}

Solution:

Given, f(x, y) = xy² --- (1)

D = {(x, y)| x >= 0, y >= 0, x² + y² <= 3}

We have to find the absolute maximum and minimum values of f on the set D.

The partial derivatives of f(x, y) is

df/dx = y²

df/dy = 2xy

Substitute y = 0 in (1)

We have x = 0

This implies that the critical point = f(0, 0) = 0.

To calculate the extreme values from x² + y² <= 3

Rewriting the expression as x² + y² = 3

y² = 3 - x²

Taking square root,

y = √(3 - x²) --- (2)

Put the value of y in (1),

f(x) = x[√(3 - x²)]²

f(x) = x(3 - x²)

f(x) = 3x - x³

On differentiating,

f’(x) = 3 - 3x²

Let f’(x) = 0

3 - 3x² = 0

3x² = 3

x² = 1

Taking square root,

x = ±1

Given, x > 0

So, the value of x is +1.

Put x = 1 in (2)

y = √(3 - 1)

y = √2

The absolute maximum is at (x, y) = (1, √2)

The absolute minimum is at (x, y) = (0, 0)

At (0, 0), f(0, 0) = 0

At (1, √2), f(1, √2) = 1(√2)² = 2

Therefore, the absolute maximum and minimum values of f are 0 and 2.

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Find the absolute maximum and minimum values of f on the set D. f(x, y)= xy², D = {(x, y)| x >= 0, y >= 0, x² + y² <= 3}

Summary:

The absolute maximum and minimum values of f on the set D. f(x, y)= xy², D = {(x, y)| x >= 0, y >= 0, x² + y² <= 3} are 0 and 2.