# find the relative maximum of f(x,y)=4xy with constraint g(x,y) = 2x+3y+xy

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You need to use the Lagrange's multipliers to evaluate the relative maximum of the given function f(x,y), subjected to the given constraint g(x,y). The Lagrange multipliers locate the critical points of extrema values.

You need to observe that f(x,y) and g(x,y) are elementary functions, hence, they have continuous partial derivatives.

You need to define a new function, such that:

` F(x,y,lambda) = f(x,y) + lambda*g(x,y)`

` F(x,y,lambda) = 4xy + lambda(2x+3y+xy)`

lambda represents a Lagrange multiplier

You need to evaluate the partial derivatives of the fuinction F(x,y,lambda), such that:

`F_x = 4y + 2lambda + lambda*y`

`F_y = 4x + 3lambda + lambda*x`

`F_lambda = 2x + 3y + xy`

Set each partial derivative equal to 0, such that:

`4y + 2lambda + lambda*y = 0 => 4y = -lambda*(2+y) => lambda = -4y/(2+y)`

`4x + 3lambda + lambda*x = 0 => 4x = -lambda(3 + x) => lambda = -4x/(3+x)`

` -4y/(2+y) = -4x/(3+x) => y/(2+y) = x/(3+x)`

`2x + 3y + xy = 0 => x(2+y) = -3y => -x/3 = y/(2+y)`

Replacing `x/(3+x) for y/(2+y)` in equation `-x/3 = y/(2+y), ` such that:

`x/(3+x) = -x/3 => -x(3+x) = 3x => x(3+x) + 3x = 0`

`x(3 + x + 3) = 0 => x(x + 6) = 0 => x = 0 => y/(2+y) = 0 => y = 0 => lambda = 0`

`x + 6 = 0 => x = -6 => y/(2+y) = 6/3 =>y/(2+y) = 2 => 4 + 2y = y => y = -4 `

` lambda = -4x/(3+x) => lambda = 24/(-3) => lambda = -8`

**Hence, evaluating the relative maximum point, under the given condition, yields (-6,-4).**

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