Respuesta :
[tex]\begin{gathered} \text{Given} \\ p(d)=-5d^2+200d-820 \\ \text{Find }d\text{ such that }p(d)=1000 \end{gathered}[/tex]
Substitute p(d) = 1000, and equate it to the given
[tex]\begin{gathered} p(d)=-5d^2+200d-820 \\ 1000=-5d^2+200d-820 \end{gathered}[/tex]Subtract both sides by 1000, and we can now solve this as a quadratic equation.
[tex]\begin{gathered} 1000-1000=-5d^2+200d-820-1000 \\ -5d^2+200d-1820=0 \end{gathered}[/tex]The quadratic formula is defined as
[tex]\begin{gathered} d=\frac{ -b \pm\sqrt{b^2 - 4ac}}{ 2a } \\ \text{where} \\ a,b,\text{ and }c\text{ are the coefficients of the standard form} \\ ax^2+bx+c \\ \\ \text{The equation }-5d^2+200d-1820\text{ has the coefficients} \\ a=-5 \\ b=200 \\ c=-1820 \end{gathered}[/tex]Substitute the following values to the quadratic formula and we get
[tex]\begin{gathered} d=\frac{ -b \pm\sqrt{b^2 - 4ac}}{ 2a } \\ d=\frac{ -200 \pm\sqrt{200^2 - 4(-5)(-1820)}}{ 2(-5) } \\ d=\frac{ -200 \pm\sqrt{40000 - 36400}}{ -10 } \\ d=\frac{ -200 \pm\sqrt{3600}}{ -10 } \\ d=\frac{ -200 \pm60\, }{ -10 } \\ \\ d_1=\frac{-200+60\, }{-10} \\ d_1=\frac{-140}{-10} \\ d_1=14 \\ \\ d_2=\frac{-200-60\, }{-10} \\ d_2=\frac{-260}{-10} \\ d_2=26 \end{gathered}[/tex]Checking
[tex]\begin{gathered} p(d)=-5d^2+200d-820 \\ \\ \text{IF }d=14 \\ p(d)=-5d^2+200d-820 \\ p(14)=-5(14)^2+200(14)-820 \\ p(14)=-5(196)+2800-820 \\ p(14)=-980+2800-820 \\ p(14)=1000 \\ \\ \text{IF }d=26 \\ p(d)=-5d^2+200d-820 \\ p(26)=-5(26)^2+200(26)-820 \\ p(26)=-5(676)+5200-820 \\ p(26)=-3380+5200-820 \\ p(26)=1000 \end{gathered}[/tex]Therefore, the prices at which the theater would earn $1,000 in profit from the comedy show each weekend is at $14, and $26.
