# Thread: partial derivative and finding an equation

1. ## partial derivative and finding an equation

Let P(x,y)= (xy)/(ax+by)
where P represents profit: x,y>0 represent sales of products X, and Y
a,b are constants>0
show that marginal profit when x=y equal to 1/(a+b)
what does it mean when x=y? finding partial deriv and set them = to each other doesnt work does it?
dp/dx =[ y(ax+by)- a(xy)]/ [(ax+by)^2]
dp/dy =[ x(ax+by) -b(xy)]/ [(ax+by)^2]

find the equation of the horizontal plane that is tangent to the graph
z= G(x,y)= x^2 -4xy - 2y^2 +12x - 12y -1
how do u do this question with 3 d graph?

2. Originally Posted by KQ_jai
Let P(x,y)= (xy)/(ax+by)
where P represents profit: x,y>0 represent sales of products X, and Y
a,b are constants>0
show that marginal profit when x=y equal to 1/(a+b)
what does it mean when x=y? finding partial deriv and set them = to each other doesnt work does it?
dp/dx =[ y(ax+by)- a(xy)]/ [(ax+by)^2]
dp/dy =[ x(ax+by) -b(xy)]/ [(ax+by)^2]
Why not? The marginal profit if given by $dP=\frac{\partial P}{\partial x}dx+ \partial P}{\partial y}dy$
if y= x, [x(ax+bx)- ax^2]/(ax+bx)^2= [ax^2+ bx^2- ax^2]/(a+b)^2x^2= b/(a+b)^2 and [x(ax+by)- bxy]/(ax+by)^2= a/(a+b)^2 so dP= (b+ a)/(a+b)^2= 1/(a+b)dx. Your definition of "marginal profit" might be that 1/(a+b).

find the equation of the horizontal plane that is tangent to the graph
z= G(x,y)= x^2 -4xy - 2y^2 +12x - 12y -1
how do u do this question with 3 d graph?

At what point? the tangent plane is different at each point. The gradient $\nabla (G- z)= (2x-4y+ 12)\vec{i}- (4x- 4y- 12)\vec{j}- \vec{k}$ is perpendicular to the surface at (x, y, z). Do you know how to find the plane with a given normal vector and point?