relative extrema for 9x3+y3/3-4xy

Question

9x^3 + y^3 / 3 - 4xy

Answer 1

Let z=(9x³+y³)/(3-4xy). This is a surface so, to find extrema, we need to work out gradients in two perpendicular directions (x direction and y direction). To do this, we need to work out two partial derivatives ∂z/∂x and ∂z/∂y, both of which must evaluate as zero, indicating a change in direction of the curved surface in both directions.

∂z/∂x=[(3-4xy)(27x²)-(9x³+y³)(-4y)]/(3-4xy)² (treat y as a constant).

When ∂z/∂x=0, (3-4xy)(27x²)-(9x³+y³)(-4y)=0,

81x²-108x³y+36x³y+4y⁴=0,

(1) 81x²-72x³y+4y⁴=0.

∂z/∂y=[(3-4xy)(3y²)-(9x³+y³)(-4x)]/(3-4xy)²=0 (treat x as a constant).

9y²-12xy³+36x⁴+4xy³=0,

(2) 9y²-8xy³+36x⁴=0.

We have two equations and two unknowns so we can find x and y to satisfy both (1) and (2).

By plotting these two equations we can see where the graphs intersect.

Let v=xy, then y=v/x.

(1)→(3): 81x²-72vx²+4v⁴/x⁴=0, 81x⁶-72vx⁶+4v⁴=0, -9x⁶(8v-9)+4v⁴=0, x³=⅔v²/√(8v-9).

(2)→(4): 9v²/x²-8v³/x²+36x⁴=0, 9v²-8v³+36x⁶=0, -v²(8v-9)+36x⁶=0, x³=v√(8v-9)/6.

Equating (3) and (4): ⅔v²/√(8v-9)=v√(8v-9)/6, 4v=8v-9, v=9/4⇒x³=⅙(9/4)√9=9/8, x=∛9/2⇒y=v/x=(9/2)/9^⅓=9^⅔/2=∛81/2.

There are two intersections: (∛9/2,∛81/2), (-∛9/2,-∛81/2). ∛9/2=1.04004 approx, ∛81/2=2.16337 approx.

The intersections are confirmed graphically.

Since 9=3² and 81=3⁴, the intersections can be written (3^⅔/2,3^1⅓/2), (-3^⅔/2,-3^1⅓/2).

These produce respectively z=-27/8 and 27/8, by plugging these values into (9x³+y³)/(3-4xy)=±(81/8+81/8)/(-6)=∓27/8. Note that 27/8=(3/2)³=3.375.

Nearby points would be (1,2): so z=(9+8)/(-5)=-3.4<-3.375 making (∛9/2,∛81/2) a maximum;  and (-1,-2): so z=-17/(-5)=3.4>3.375 making (-∛9/2,-∛81/2) a minimum.