Lagrange multiplier question(Please help)

Question

Show that the function f(x,y) = 8(x^2)y, subject to 3x - y = 9, does not have an absolute minimum or maximum.

(hint: Solve the constraint for y and substitute into f.)

Hmm. even though there is a hint, I do not get it since the textbook has never mentioned about this type of

question in any of its previous pages. Please show me your working thanks.

Answer 1

Problem: Minimize f(x,y) = 8(x^2)y, subject to 3x - y = 9

Let m be a Lagrange Multiplier, and consider the Lagrangian:

L(x,y,m) = f(x,y) - m(3x - y -9) = 8yx² - m(3x - y -9)

Equate the Partial Derivatives to zero:

L_x = 16yx - 3m = 0

L_y = 8x² -m(-1) = 0

L_m = 3x - y - 9 = 0

Rearranging,

3m = 16xy

m = -8x²

Taking the ratio yields:

3 = -2y/x, or

3x = -2y.

The Constraint can now be used to get values for x, y:

3x - y = 9

-2y - y = 9

y = -3

3x = -2y = -2(-3)

x = 6

Let's now take the 2nd Partial Derivatives:

L_x = 16yx - 3m = 16xy - 3(-8x²) = 16xy + 24x²

L_xx = 16y + 48x = 16(-3) + 48(6) = 16*3(-1 +3*2) = 16*3*5 = 48*5 = 240

L_xy = 16x = 16(6) = 96

L_y = 8x² -m(-1) = L_y = 8x² + -8x² = 0

L_yy = 0

L_yx = 0

The Hessian Matrix is zero, so no conclusion about (6,-3) is possible.

Caution: The Hessian Matrix I computed was based on L, but I could be wrong - perhaps it should be based on f(x,y) instead.