Question

find maxima and minima of the function f(x,y)=x^3y^2(6-x-y)​

Answer 1

Step-by-step explanation:

(xy)=x³,y²(6-x-y)

(xy)=x³+x,y²+y+6

(xy)=x⁴,y³+6

x-x⁴, y-y³=6

-x³,-y²=6

x³,y²=6

3x,2y=6

x+y=6

5xy=6

xy=6/5

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Answer 2

The maxima and minima of the given function is

128 and -27 respectively.

Step by step explanation:

Maxima and minima of the function f(x,y)=x^3y^2(6-x-y) can be determined by taking the partial derivatives of the function with respect to x and y, and equating them to zero.

Letting fx=0 and fy=0, we get

fx= 3x^2y^2(6-x-y) – x^3y^2 = 0

fy= 2x^3y(6-x-y) – x^3y^2 = 0

Solving these two equations, we get the maxima and minima of the function at x=2, y=2 and x=3, y=1.

Therefore, the maxima and minima of the function f(x,y)=x^3y^2(6-x-y) is (2,2) and (3,1) respectively.

Further solving the equation we get,

Maxima:

When x=2 and y=2, f(x,y)=128, which is the absolute maximum value of the function.

Minima:

When x=3 and y=1, f(x,y)=-27, which is the absolute minimum value of the function.

To know more about partial derivatives, click the link below

https://brainly.in/question/9360534

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