Question

The function f (x comma y )equals 3 xy has an absolute maximum value and absolute minimum value subject to the constraint 3 x squared plus 3 y squared minus 5 xyequals121. Use Lagrange multipliers to find these values.

Answer 1

Answer:

The maximum value of f is 363, which is reached in (11,11) and (-11,-11) and the minimum value of f is -33, which is reached in (√11,-√11) and (-√11,√11)

Step-by-step explanation:

f(x,y) = 3xy, lets find the gradient of f. First lets compute the derivate of f in terms of x, thinking of y like a constant.

$f_x(x,y) = 3y$

In a similar way

$f_y(x,y) = 3x$

Thus,

$\nabla{f} = (3y,3x)$

The restriction is given by g(x,y) = 121, with g(x,y) = 3x²+3y²-5xy. The partial derivates of g are

[ŧex] g_x(x,y) = 6x-5y [/tex]

$g_y(x,y) = 6y - 5x$

Thus,

$\nabla g(x,y) = (6x-5y,6y-5x)$

For the Langrange multipliers theorem, we have that for an extreme (x0,y0) with the restriction g(x,y) = 121, we have that for certain λ,

• $f_x(x_0,y_0) = \lambda \, g_x(x0,y0)$
• $f_y(x_0,y_0) = \lambda \, g_y(x_0,y_0)$
• $g(x_0,y_0) = 121$

This can be translated into

• $3y = \lambda (6x-5y)$
• $3x = \lambda (-5x+6y)$
• $3 (x_0)^2 + 3(y_0)^2 - 5\,x_0y_0 = 121$

If we sum the first two expressions, we obtain

$3x + 3y = \lambda (x+y)$

Thus, x = -y or λ=3.

If x were -y, then we can replace x for -y in both equations

3y = -11 λ y

-3y = 11 λ y, and therefore

y = 0, or λ = -3/11.

Note that y cant take the value 0 because, since x = -y, we have that x = y = y, and g(x,y) = 0. Therefore, equation 3 wouldnt hold.

Now, lets suppose that λ=3, if that is the case, we can replace in the first 2 equations obtaining

• 3y = 3(6x-5y) = 18x -15y

thus, 18y = 18x

y = x

and also,

• 3x = 3(6y-5x) = 18y-15x

18x = 18y

x = y

Therefore, x = y or x = -y.

If x = -y:

Lets evaluate g in (-y,y) and try to find y

g(-y,y) = 3(-y)² + 3y*2 - 5(-y)y = 11y² = 121

Therefore,

y² = 121/11 = 11

y = √11 or y = -√11

The candidates to extremes are, as a result (√11,-√11), (-√11, √11). In both cases, f(x,y) = 3 √11 (-√11) = -33

If x = y:

g(y,y) = 3y²+3y²-5y² = y² = 121, then y = 11 or y = -11

In both cases f(11,11) = f(-11,-11) = 363.

We conclude that the maximum value of f is 363, which is reached in (11,11) and (-11,-11) and the minimum value of f is -33, which is reached in (√11,-√11) and (-√11,√11)