Profit

[TheNextOne]

A company manufactures three kinds of products A, B, and C, the
total number of products being 900 pieces. The profit resulting from producing x
hundred units of the product A, y hundred units of B, and z hundred units of C is
P = xy + xz + yz dollars. How many units of each of the products A, B, and C must be manufactured in order to maximize the profit.

I did the following:

xy + xz + yz - q(x+y+z-900)= F(x,y,z)

Fx (x,y,z)= y+z-q
Fy (x,y,z)= x+z-q
Fz (x,y,z)= y+x-q

I cant seem to solve the system. to find the values.

 

[galactus]

Use the function given and x+y+z=900

\(\displaystyle z=900-x-y\)

So, \(\displaystyle P=xy+x(900-x-y)+y(900-x-y)\)

Differentiate:

\(\displaystyle \frac{\partial{P}}{\partial{x}}=-2x-y+900\)

\(\displaystyle \frac{\partial{P}}{\partial{y}}=-2y-x+900\)

the coordinates of the critical points of P satisfy:

\(\displaystyle {-}2x-y+900=0\) and \(\displaystyle {-}2y-x+900=0\)

Solve the first one for y and sub into the second, solve for x.

You can use the second partials test to see if you found any relative extrema.