Question

find the maximum and minimum values of x^3y^2-x^4y^2-x^3y^3



hi Mayank ​

Answer 1

Explanation AND 2

Maxima and minima of multivariable function f(x, y) = 6x³y² - x^² - x³y³

multivariable-calculus optimization

f(x, y) = 6x³y² - xy² - x³y³

8x = 18x²y² - 4x³y² - 3x²y³

sf dy 12x³y – 2x¹y − 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

A1 sf sf : 36xy² — 12x²y² – 6xy³ 12x³ 2x² - 6x³y

sf dxdy of dydx 36x²y — 8x³y – 9x²y² -

0² f

მ2

8² f

dx dy

A2

=

8² f მყმე: 8² f მ/2

After plugging in the point (3,2) we get A₁ <0 and A2 0, so (3,2) is maxima. Now then I try to plug in (x,0) and (0,y) I obviously get A1 = 0 and A₂2 = 0 and I can't tell, using Sylvester's criterion, if those points are minima or maxima or neither. What should I do?

Answer 2

Explanation:

Note that the function f(x, x) = 6x5 − 2x6 is zero at (3, 3) , which is a point on the "zero-line" x + y = 6 . It has a maximum at (52, 52) , along a "ridge" formed in the surface, which includes the relative maximum you found nearby at (3, 2) . ]

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