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March 31, 2015

March 31, 2015

Posted by **Anonymous** on Sunday, June 23, 2013 at 9:42am.

{2x2−6xy+2y2+43x+43yx2+y2+5x+5y==174,30.

Find the largest possible value of |xy

- maths -
**MathMate**, Sunday, June 23, 2013 at 12:50pmThe problem we have on hand is:

Maximize |xy|

subject to:

f(x,y)=0

where

f(x,y)=2y^2+43x^2y-6xy+5y+2x^2+48x-174.3

To solve the problem, we will solve for the extreme values of xy, and choose from the extreme values the largest value of |xy|.

To do this, we would use Lagrange multipliers:

Objective function,

Z(x,y)=xy+L(f(x,y))

constraint:

f(x,y)

Set up equations by equating the partial derivatives with respect to x, y and L (the Lagrange multiplier) to zero.0

Z_{x}(x,y)=y+L(86xy-6y+4x+48) = 0

Z_{y}(x,y)=x+L(4y+43x^2-6x+5) = 0

Z_{L}(x,y)=2y^2+43x^2*y-6xy+5*y+2x^2+48x-174.3=0

We now need to solve the algebraic system for x and y by first eliminating L from the first two equations to get the following:

g(x,y)=x(86xy-6y+4x+48)-y(4y+43x^2-6x+5)=0

Together with

f(x)=2y^2+43x^2y-6xy+5y+2x^2+48x-174.3 = 0

we obtain an algebraic system from which we need to extract the real roots.

This can be obtained by using Newton's method (in two variables) with starting points (-0.8,4) and (0.6,5) by substituting trial integer values of y from 1-5.

The solutions are

(-0.8491, 4.2984) giving xy=-3.6500

and

(0.6472, 4.9170) giving xy=3.1821

So the largest possible value of |xy|=3.64997 (approximately)

It is possible that the solution be obtained analytically (not numerically) using some skilled manipulations.

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