Question

A manager estimates a band's profit p for a concert by using the function p(t)=-200t^2+2500t-c, where t is the price per ticket and c is the band's op

Quadratic function and equation
ANSWERED
asked 2020-12-02

A manager estimates a band's profit p for a concert by using the function \(p(t)=-200t^2+2500t-c,\), where t is the price per ticket and c is the band's operation cost. The table shows the band's operating cost at three different concert locations. What range of ticket prices should the band charge at each location in order to make a profit of at least $1000 at each concert?
Band's Costs
Location Operating Cost
Freemont Park $900
Saltillo Plaza $1500
Riverside Walk $2500

Answers (1)

2020-12-03
freemont park: c=900
p(t) = 1000 subsitute variables into equation
\(\displaystyle{p}{\left({t}\right)}=-{200}{\left({t}^{{2}}\right)}+{2500}{t}-{c}\)
\(\displaystyle{\left({1000}\right)}=-{200}{\left({t}^{{2}}\right)}+{2500}{t}-{\left({900}\right)}\)
\(\displaystyle{0}=-{200}{\left({t}^{{2}}\right)}+{2500}{t}-{1900}\)
\(\displaystyle{200}{t}^{{2}}-{2500}{t}+{1900}={0}\)
a = 200, b = -2500 c = 1900 quadratic equation
\(\displaystyle{t}=\frac{{{2500}\pm\sqrt{{{2500}^{{2}}-{4}{\left({200}\right)}{\left({1900}\right)}}}}}{{2}}{\left({200}\right)}\)
\(\displaystyle{t}=\frac{{{2500}\pm\sqrt{{{4730000}}}}}{{400}}\)
\(\displaystyle{t}=\frac{{25}}{{4}}+\frac{\sqrt{{{473}}}}{{4}}\)
\(\displaystyle{t}=\frac{{25}}{{4}}-\frac{\sqrt{{{473}}}}{{4}}\)
\(\displaystyle{t}={6.25}+{5.43714079273}=\${11.69}\) or
\(\displaystyle{t}={6.25}-{5.43714079273}=\${0.81}\)
saltillo plaza: c = 1500
p(t) = 1000 subsitute variables into equation
\(\displaystyle{p}{\left({t}\right)}=-{200}{\left({t}^{{2}}\right)}+{2500}{t}-{c}\)
\(\displaystyle{\left({1000}\right)}=-{200}{\left({t}^{{2}}\right)}+{2500}{t}-{\left({1500}\right)}\)
\(\displaystyle{0}=-{200}{\left({t}^{{2}}\right)}+{2500}{t}-{2500}\)
\(\displaystyle{200}{t}^{{2}}-{2500}{t}+{2500}={0}\)
a = 200, b = -2500 c = 2500 quadratic equation
\(\displaystyle{t}=\frac{{{2500}\pm\sqrt{{{2500}^{{2}}-{4}{\left({200}\right)}{\left({2500}\right)}}}}}{{2}}{\left({200}\right)}\)
\(\displaystyle{t}=\frac{{{2500}\pm\sqrt{{{4250000}}}}}{{400}}\)
\(\displaystyle{t}=\frac{{25}}{{4}}+{\left(\frac{{5}}{{4}}\right)}\sqrt{{{17}}}=\${11.40}\) or
\(\displaystyle{t}=\frac{{25}}{{4}}-{\left(\frac{{5}}{{4}}\right)}\sqrt{{{17}}}=\${1.09}\)
riverside walk: c = 2500
p(t) = 1000 subsitute variables into equation
\(\displaystyle{p}{\left({t}\right)}=-{200}{\left({t}^{{2}}\right)}+{2500}{t}-{c}\)
\(\displaystyle{\left({1000}\right)}=-{200}{\left({t}^{{2}}\right)}+{2500}{t}-{\left({2500}\right)}\)
\(\displaystyle{0}=-{200}{\left({t}^{{2}}\right)}+{2500}{t}-{3500}\)
\(\displaystyle{200}{t}^{{2}}-{2500}{t}+{3500}={0}\)
a = 200, b = -2500 c = 3500 quadratic equation
\(\displaystyle{t}=\frac{{{2500}\pm\sqrt{{{2500}^{{2}}-{4}{\left({200}\right)}{\left({3500}\right)}}}}}{{2}}{\left({200}\right)}\)
\(\displaystyle{t}=\frac{{{2500}\pm\sqrt{{{3450000}}}}}{{400}}\)
t = $10.89 or
t = $1.61
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