Question

find the
f(x, y, z)
maximum value of the function
(3-2x2 -- 2y2) where 3xy - 2+7=0​

Answer 1

The maximum value of the function f(x, y, z) is 3/2.

Given,

3xy - 2 + 7 = 0​

To find,

f(x, y, z)

Solution,

To find the maximum value of the function f(x, y, z), we need to take the partial derivatives of f with respect to each of the variables x, y, and z, and then set those partial derivatives equal to 0.

With respect to x

df/dx = -4x(3 - 2x² - 2y²)

with respect to y is:

df/dy = -4y(3 - 2x² - 2y²)

Since there is no z variable in the function f, the partial derivative of f with respect to z is 0.

To find the maximum value of the function, we set each of the partial derivatives equal to 0 and solve for x and y:

-4x(3 - 2x² - 2y²) = 0

3 - 2x² - 2y² = 0

2x² + 2y² = 3

x² + y² = 3/2

-4y(3 - 2x² - 2y²) = 0

3 - 2x² - 2y² = 0

2x² + 2y² = 3

x² + y² = 3/2

Since the equation x² + y² = 3/2 represents a circle with radius √(3/2), the maximum value of the function f(x, y, z) occurs at the point

(x, y) = √(3/2), 0.

f(√(3/2), 0, z)

f= 3 - 2(3/2)

f= 3/2.

Therefore, the maximum value of the function f(x, y, z) is 3/2.

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