Question

Find max(x+3y) if: x^2+xy+4y^2<(or equal)3

Answer 1

Answer:

Step-by-step explanation:

Begin with x2 + xy + y2 = 1 . Differentiate both sides of the equation, getting

D ( x2 + xy + y2 ) = D ( 1 ) ,

2x + ( xy' + (1)y ) + 2 y y' = 0 ,

so that (Now solve for y' .)

xy' + 2 y y' = - 2x - y ,

(Factor out y' .)

y' [ x + 2y ] = - 2 x - y ,

and the first derivative as a function of x and y is

(Equation 1)

$ y' = \displaystyle{ - 2 x - y \over x + 2y } $ .

To find y'' , differentiate both sides of this equation, getting

$ y'' = \displaystyle{ (x + 2y) D(-2x - y) - (-2x - y) D(x + 2y) \over (x + 2y)^2 } $

$ = \displaystyle{ (x + 2y)(-2 - y') - (-1)(2x + y) (1 + 2y') \over (x + 2y)^2 } $

$ = \displaystyle{ -2x - 4y -xy' - 2yy' + 2x + y + 4xy' + 2yy' \over (x + 2y)^2 } $

$ = \displaystyle{ 3xy' - 3y \over (x + 2y)^2 } $ .

Use Equation 1 to substitute for y' , getting

$ y'' = \displaystyle{ 3x \Big( \displaystyle{ - 2 x - y \over x + 2y } \Big) - 3y \over (x + 2y)^2 } $

(Get a common denominator in the numerator and simplify the expression.)

$ = \displaystyle{ 3x \Big( \displaystyle{ -2x-y \over x+2y } \Big)

- 3y \Big\{ \displaystyle{ x+2y \over x+2y } \Big\} \over (x + 2y)^2 } $

$ = \displaystyle{ \displaystyle{ 3x (-2x-y) - 3y(x+2y) \over x+2y } \over

\displaystyle{ (x + 2y)^2 \over 1 } } $

$ = \displaystyle{ { -6x^2 - 3xy - 3xy - 6y^2 \over x+2y }

{ 1 \over (x + 2y)^2 } } $

$ = \displaystyle{ -6x^2 - 6y^2 - 6xy \over (x+2y)^3 } $ .

This answer can be simplified even further. Note that the original equation is

x2 + xy + y2 = 1 ,

so that

(Equation 2)

x2 + y2 = 1 - xy .

Use Equation 2 to substitute into the equation for y'' , getting

$ y'' = \displaystyle{ -6x^2 - 6y^2 -6xy \over (x+2y)^3 } $

$ = \displaystyle{ -6 (x^2 + y^2) - 6xy \over (x+2y)^3 } $

$ = \displaystyle{ -6 (1 - xy) - 6xy \over (x+2y)^3 } $

$ = \displaystyle{ -6 + 6xy - 6xy \over (x+2y)^3 } $ ,

and the second derivative as a function of x and y is

$ y'' = \displaystyle{ -6 \over (x+2y)^3 } $ .