Question

Discuss the maximum or minimum values of u, if u=ax^3y^2-x^4y^2-x^3y^3

Answer 1

Answer:

Δ1=0 and Δ2=0

Step-by-step explanation:

Points, in which partial derivatives ar equal to 0 are: (3,2), (x,0), (0,y), x,y are any real numbers. Now I find second derivatives

Let f(x) be a function. Find the first derivative of f(x), which is f'(x).

Equate the first derivative f'(x) to zero and solve for x, which are called critical numbers.

Find the second derivative of f(x), which is f"(x).

If f"(x) < 0 for some value of x, say x = a, then the function f(x) is maximum at x = a.

If f"(x) > 0 for some value of x, say x = b, then the function f(x) is minimum at x = b.

To get maximum and minimum values of the function substitute x = a and x = b in f(x).

Maximum value = f(a)

Minimum value = f(b)

f(x, y) = x3 y2 ( 6 − x − y )  ,

fx = x2 y2 ( 18 − 4x − 3y )  ,  

fy = x3 y ( 12 − 2x − 3y )  ,

fxx = 6 x y2 ( 6 − 2x − y )  ,  

fyy = 2 x3 ( 6 − x − 3y )  ,  and

fxy = x2 y ( 36 − 8x − 9y )  .

f(x,y)=6x3y2−x4y2−x3y3

δfδx=18x2y2−4x3y2−3x2y3

δfδy=12x3y−2x4y−3x3y2

Points, in which partial derivatives ar equal to 0 are: (3,2), (x,0), (0,y), x,y are any real numbers. Now I find second derivatives

After plugging in the point (3,2) we get Δ1<0 and Δ2>0, so (3,2) is maxima. Now then I try to plug in (x,0) and (0,y) I obviously get Δ1=0 and Δ2=0

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