Question

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 + 8y
​C(x,y) =x^2 − 3xy + 8y^2 + 14x − 50y − 4

Determine 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.

Answer 1

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.

$P(x,y)=R(x,y)-C(x,y)\\ P(x,y)=6x+8y-x^{2}+3xy-8y^{2} -14x+50y+4\\ P(x,y)=-8x+58y-x^{2} -8y^{2} +3xy+4$

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:

$P_{x}=-8-2x+3y=0\\  P_{y} =58-16y+3x=0$

So, let's solve the equation system:

First, isolate x:

Eq. 1 $2x=3y-8$

Eq. 2$3x=16y-58$

Multiply equation 1 by (-3) and equation 2 by 2:

$-6x=-9y+24\\ 6x=32y-116$

Sum the equations:

$0=23y-92\\ y=\frac{92}{23}=4$

Find x with eq. 1 or 2:

$x=\frac{3y-8}{2}= \frac{3*4-8}{2}=2$