A community theater uses the function p(d)=-4d^2+200d-100 to model the profit (in dollars) expected in a weekend when the tickets to a comedy show are priced at d dollars each a. Write and solve an equation to find out the grices at which the theater would ears 1.500 in profit from the comedy show each weekend. Type your response in the space below. Equation: -4d^2+200d-100=0 Solutions: d=25 d=35 b. At what price would the theater make the maximum profit? Type your answer in the box below. K25 What is the masinun profit? Type your anywer in the bew below. 52.500 Show your reasoning. Type your response in the space below.

a. To find the prices at which the theater would earn $1,500 in profit from the comedy show each weekend, we need to solve the equation $-4d^{2}+200d-100=1500$.<br /><br />First, let's rewrite the equation as $-4d^{2}+200d-1600=0$.<br /><br />Next, we can solve this quadratic equation by factoring or using the quadratic formula. In this case, we will use the quadratic formula:<br /><br />$d=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$<br /><br />where $a=-4$, $b=200$, and $c=-1600$.<br /><br />Plugging in these values, we get:<br /><br />$d=\frac{-200\pm\sqrt{200^2-4(-4)(-1600)}}{2(-4)}$<br /><br />Simplifying further, we have:<br /><br />$d=\frac{-200\pm\sqrt{40000-25600}}{-8}$<br /><br />$d=\frac{-200\pm\sqrt{14400}}{-8}$<br /><br />$d=\frac{-200\pm120}{-8}$<br /><br />So, the solutions are $d=25$ and $d=35$.<br /><br />b. To find the price at which the theater would make the maximum profit, we need to find the vertex of the parabola represented by the function $p(d)=-4d^{2}+200d-100$.<br /><br />The x-coordinate of the vertex of a parabola in the form $y=ax^2+bx+c$ is given by $-\frac{b}{2a}$.<br /><br />In this case, $a=-4$ and $b=200$, so the x-coordinate of the vertex is $-\frac{200}{2(-4)}=25$.<br /><br />Therefore, the theater would make the maximum profit when the price is $25.<br /><br />c. To find the maximum profit, we can substitute the x-coordinate of the vertex ($d=25$) into the function $p(d)=-4d^{2}+200d-100$.<br /><br />$p(25)=-4(25)^{2}+200(25)-100$<br /><br />$p(25)=-4(625)+5000-100$<br /><br />$p(25)=-2500+5000-100$<br /><br />$p(25)=2500$<br /><br />Therefore, the maximum profit is $2,500.