MATH SOLVE

4 months ago

Q:
# A company produces two types of solar panels per year: x thousand of type A and y thousand of type B. The revenue and cost equations, in millions of dollars, for the year are given as follows.R(x,y) = 6x + 8yC(x,y) =x^2 − 3xy + 8y^2 + 14x − 50y − 4Determine how many of each type of solar panel should be produced per year to maximize profit. The company will achieve a maximum profit by selling nothing solar panels of type A and selling nothing solar panels of type B.

Accepted Solution

A:

Answer:x=2, y=4. 2 thousand of A panels and 4 of B.Step-by-step explanation:First, the profit is determined by the revenue minus the cost, so built a profit equation with that information.[tex]P(x,y)=R(x,y)-C(x,y)\\ P(x,y)=6x+8y-x^{2}+3xy-8y^{2} -14x+50y+4\\ P(x,y)=-8x+58y-x^{2} -8y^{2} +3xy+4[/tex]Then, use the partial derivative criteria to determine which is the maximum.The partial derivative criteria says that in the local maximum or minimum, the partial derivatives are equal to zero, so:[tex]P_{x}=-8-2x+3y=0\\ P_{y} =58-16y+3x=0[/tex]So, let's solve the equation system:First, isolate x: Eq. 1 [tex]2x=3y-8[/tex] Eq. 2[tex]3x=16y-58[/tex]Multiply equation 1 by (-3) and equation 2 by 2:[tex]-6x=-9y+24\\ 6x=32y-116[/tex]Sum the equations:[tex]0=23y-92\\ y=\frac{92}{23}=4[/tex]Find x with eq. 1 or 2:[tex]x=\frac{3y-8}{2}= \frac{3*4-8}{2}=2[/tex]