A company manufactures 2 models of MP3 players. Let x represent the number (in m

Question

A company manufactures 2 models of MP3 players. Let x represent the number (in millions) of the first model made, and let y represent the number (in millions) of the second model made. The company's revenue can be modeled by the equation R(x, y) = 130x + 140y - 3x^2 - 2y^2 - xy Find the marginal revenue equations R_x (x, y) = R_y (x, y) = We can achieve maximum revenue when both partial derivatives are equal to zero. Set R_x = 0 and R_y = 0 and solve as a system of equations to the find the production levels that will maximize revenue. Revenue will be maximized when: x = y =

Explanation / Answer

given R(x,y)=130x+140y -3x2-2y2-xy

Rx(x,y)=130+0 -6x-0-y

Rx(x,y)=130 -6x-y

Ry(x,y)=0+140 -0-4y-x

Ry(x,y)=140-4y-x

Rx(x,y)=0,Ry(x,y)=0

130 -6x-y=0 ,140-4y-x=0

=> y=130-6x ,x=140-4y

=>x=140-4(130-6x)

=>x=140-520+24x

=>23x=380

=>x=(380/23)

y=130-6x,x=380/23

=>y=130-6(380/23)

=>y=(710/23)

Rx(x,y)=130 -6x-y,Ry(x,y)=140-4y-x

Rxx(x,y)= -6,Ryy(x,y)=-4,Rxy(x,y)=-1,D=(Rxx*Ryy)-(Rxy)2

D=(-6*-4)-(-1)2=23>0 ,Rxx(x,y)<0

so revenue is maximised when

x=(380/23)

y=(710/23)

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