Question

a community theater uses the function p(d)=-5d^2+200d-820 to model the profit (in dollars) expected in a weekend when the tickets to a comedy show are priced at d dollars each. write and solve an equation to find out the prices at which the theater would earn $1,000 in profit from the comedy show each weekend.show your work

Answer 1

`\begin{gathered} \text{Given} \\ p(d)=-5d^2+200d-820 \\ \text{Find }d\text{ such that }p(d)=1000 \end{gathered}`

Substitute p(d) = 1000, and equate it to the given

`\begin{gathered} p(d)=-5d^2+200d-820 \\ 1000=-5d^2+200d-820 \end{gathered}`

Subtract both sides by 1000, and we can now solve this as a quadratic equation.

`\begin{gathered} 1000-1000=-5d^2+200d-820-1000 \\ -5d^2+200d-1820=0 \end{gathered}`

The quadratic formula is defined as

`\begin{gathered} d=( -b \pm√(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}`

Substitute the following values to the quadratic formula and we get

`\begin{gathered} d=( -b \pm√(b^2 - 4ac))/( 2a ) \\ d=( -200 \pm√(200^2 - 4(-5)(-1820)))/( 2(-5) ) \\ d=( -200 \pm√(40000 - 36400))/( -10 ) \\ d=( -200 \pm√(3600))/( -10 ) \\ d=( -200 \pm60\, )/( -10 ) \\ \\ d_1=(-200+60\, )/(-10) \\ d_1=(-140)/(-10) \\ d_1=14 \\ \\ d_2=(-200-60\, )/(-10) \\ d_2=(-260)/(-10) \\ d_2=26 \end{gathered}`

Checking

`\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}`

Therefore, the prices at which the theater would earn $1,000 in profit from the comedy show each weekend is at $14, and $26.