Posted by Anonymous on .
The real numbers x and y satisfy the nonlinear system of equations
{2x2−6xy+2y2+43x+43yx2+y2+5x+5y==174,30.
Find the largest possible value of xy

maths 
MathMate,
The problem we have on hand is:
Maximize xy
subject to:
f(x,y)=0
where
f(x,y)=2y^2+43x^2y6xy+5y+2x^2+48x174.3
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(86xy6y+4x+48) = 0
Z_{y}(x,y)=x+L(4y+43x^26x+5) = 0
Z_{L}(x,y)=2y^2+43x^2*y6xy+5*y+2x^2+48x174.3=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(86xy6y+4x+48)y(4y+43x^26x+5)=0
Together with
f(x)=2y^2+43x^2y6xy+5y+2x^2+48x174.3 = 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 15.
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.