Respuesta :
Using partial derivatives, we know that in order to optimize profit, the company should produce 7,000 type A solar panels and 8,000 type B solar panels annually.
What are partial derivatives?
A function of two or more variables can yield partial derivatives.
Writing the derivative with regard to x while treating all other variables as constants allows one to determine the partial derivative of a function with respect to one variable, let's say x.
The partial derivatives for each of the other
So, for a business making two different types of solar panels, the revenue and cost functions are provided by:
[tex]\begin{aligned}&R(x, y)=4 x+2 y \\&C(x, y)=x^2-4 x y+9 y^2+22 x-114 y-5\end{aligned}[/tex]
If revenue and cost are stated in millions of dollars annually, x represents the thousands of type A solar panels produced and sold annually, and y represents the thousands of type B solar panels produced and sold annually.
By deducting costs from revenues, the profit function is obtained as follows:
[tex]\begin{aligned}&P(x, y)=R(x, y)-C(x, y) \\&P(x, y)=4 x+2 y-\left(x^2-4 x y+9 y^2+22 x-114 y-5\right) \\&P(x, y)=-x^2-9 y^2-18 x+116 y+4 x y+5\end{aligned}[/tex]
The profit function's partial derivatives with regard to x and y are as follows:
[tex]\begin{aligned}&\frac{\partial P}{\partial x}=-2 x+4 y-18 \\&\frac{\partial P}{\partial y}=4 x-18 y+116\end{aligned}[/tex]
When both of these partial derivatives are equal to zero, the profit is maximized:
[tex]\begin{aligned}&0=-2 x+4 y-18 \\&0=4 x-18 y+116\end{aligned}[/tex]
The first equation above is multiplied by two, and the two equations are then added to yield:
[tex]\begin{aligned}&0=-10 y+80 \\&y=8\end{aligned}[/tex]
Inputting into the first equation results in:
[tex]\begin{aligned}&0=-2 x+4(8)-18 \\&2 x=14 \\&x=7\end{aligned}[/tex]
Therefore, using partial derivatives, we know that in order to optimize profit, the company should produce 7,000 type A solar panels and 8,000 type B solar panels annually.
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