y" = -4/(4*x+9*y-12)-4*(-4*x-4*y+4)/(4*x+9*y-12)^2
y"= -4(5y−8)/(4x+9y−12)^2
but x+y=1 at max so
y"= 4(8-5y)/(5y-8)
if 4x^2+8xy+9y^2-8x-24y+4=0 show that when dy/dx=0, x+y=1 and d^2y/dx^2=4/(8-5y) hence find the maximum and minimum values of y
plz help so far so good i got
dy/dx=(4-4y-4x)/(4x+9y-12)
that help me to get x+y=1
but where i could not get is the second derivative of it so that i could get d/(8-5y) and then finish it up plz help me
2 answers
4x^2+8xy+9y^2-8x-24y+4=0
8xdx+8xdy+8ydx+18ydy-8dx-24dy=0
(8x+18y-24)dy= -(8x+8y-8)dx
dy/dx = (-4x-4y+4)/(4x+9y-12)agree
numerator = 0 when x+y=1 yes
find d/dx(dy/dx)of (-4x-4y+4)/(4x+9y-12)
= [(4x+9y-12)(-4dx-4dy)-(-4x-4y+4)(4dx+9dy)]/dx[(4x+9y-12)]^2
=4[(4x+9y-12)(-dx-dy)+(+x+y-1)(4dx+9dy)]/dx[(4x+9y-12)]^2
= 4[-4xdx-9ydx+12dx -4xdy-9ydy+12dy+4xdx+4ydx-4dx+9xdy+9ydy-9dy]/dx[(4x+9y-12)]^2
=4[8dx-5ydx+5xdy +3dy]/dx[(4x+9y-12)]^2
= 4[8-5y+5xdy/dx+3dy/dx]
/[(4x+9y-12)]^2
but dy/dx = 0 here and x=1-y
= 4(8-5y)/[4(1-y)+9y-12]^2
= 4(8-5y)/[-8+5y]^2
the same as
4(8-5y)/(8-5y)^2
=4/(8-5y)
whew
8xdx+8xdy+8ydx+18ydy-8dx-24dy=0
(8x+18y-24)dy= -(8x+8y-8)dx
dy/dx = (-4x-4y+4)/(4x+9y-12)agree
numerator = 0 when x+y=1 yes
find d/dx(dy/dx)of (-4x-4y+4)/(4x+9y-12)
= [(4x+9y-12)(-4dx-4dy)-(-4x-4y+4)(4dx+9dy)]/dx[(4x+9y-12)]^2
=4[(4x+9y-12)(-dx-dy)+(+x+y-1)(4dx+9dy)]/dx[(4x+9y-12)]^2
= 4[-4xdx-9ydx+12dx -4xdy-9ydy+12dy+4xdx+4ydx-4dx+9xdy+9ydy-9dy]/dx[(4x+9y-12)]^2
=4[8dx-5ydx+5xdy +3dy]/dx[(4x+9y-12)]^2
= 4[8-5y+5xdy/dx+3dy/dx]
/[(4x+9y-12)]^2
but dy/dx = 0 here and x=1-y
= 4(8-5y)/[4(1-y)+9y-12]^2
= 4(8-5y)/[-8+5y]^2
the same as
4(8-5y)/(8-5y)^2
=4/(8-5y)
whew
1 answer