Respuesta :
We have that the demand equation for a certain product is given by:
[tex]P=132-0.025x[/tex]where x is the number of units produced. Also, that the total revenue is:
[tex]R=xP[/tex]And we want to determine prices P that would yield a revenue of 7280 dollars. This means that we want to find values x such that the following equation holds:
[tex]7280=xP=x(132-0.025x)[/tex]We will solve the equation by the quadratic formula. First, we will use the distributive property and we clear out the left side of the equation, to obtain:
[tex]\begin{gathered} 7280=132x-0.025x^2 \\ 0=132x-0.025x^2-7280 \\ 0=-0.025x^2+132x-7280 \end{gathered}[/tex]In this equation:
[tex]\begin{gathered} a=-0.025 \\ b=132 \\ c=-7280 \end{gathered}[/tex]And thus, replacing on the quadratic formula we obtain:
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ =\frac{-132\pm\sqrt[]{(132)^2-4(-0.025)(-7280)}}{2(-0.025)} \\ =\frac{-132\pm\sqrt[]{17424-728}}{-0.05} \\ =\frac{-132\pm\sqrt[]{16696}}{-0.05} \\ \approx\frac{-132\pm129.21}{-0.05} \end{gathered}[/tex]Now, for the two solutions, we separate them with the signs + and -.
[tex]\begin{gathered} x_1=\frac{-132+129.21}{-0.05} \\ \approx55.8 \end{gathered}[/tex]And for the other one:
[tex]\begin{gathered} x_2=\frac{-132-129.21}{-0.05} \\ \approx5224.2 \end{gathered}[/tex]This means that the lowest of the prices should be 56 dollars, and the highest price 5224 dollars.
