Respuesta :
Answer:
Let the price for one $1 increase be
[tex]=x[/tex]Let the revenue be represented as
[tex]=R[/tex]The number of tickets sold will be represented below as
[tex]=500-10x[/tex]The price per ticket will be represented below as
[tex]=20+x[/tex]The revenue is calculated using the formula below
[tex]R=Nu\text{mber of tickets}\times price\text{ per tickets}[/tex]By substituting the values above, we will have
[tex]\begin{gathered} R=Nu\text{mber of tickets}\times price\text{ per tickets} \\ R=(500-10x)(20+x) \\ \end{gathered}[/tex]Expanding the brackets, we will have
[tex]\begin{gathered} R=(500-10x)(20+x) \\ R=500(20+x)-10x(20+x) \\ R=10000+500x-200x-10x^2 \\ R=10000+300x-10x^2 \end{gathered}[/tex]Step 1;
We would calculate the value of x that will give maximum revenue
We will use the expression below
[tex]x=-\frac{b}{2a}[/tex]The general form of a quadratic equation is given below as
[tex]\begin{gathered} R=-10x^2+300x+10000 \\ y=ax^2+bx+c \\ By\text{ comparing coefficents, we will have} \\ a=-10,b=300,c=10000 \end{gathered}[/tex]By substituting the values, we will have
[tex]\begin{gathered} x=-\frac{b}{2a} \\ x=-\frac{300}{(2\times-10)} \\ x=\frac{-300}{-20} \\ x=15 \end{gathered}[/tex]Therefore,
The price per ticket that will maximize profit will be
[tex]\begin{gathered} =20+x \\ =20+15 \\ =35 \end{gathered}[/tex]Hence,
The price per ticket that will maximize profit will be = $35
Step 2:
To calculate the maximum revenue, we will substitute the value of x= 15 in the equation for the revenue given below
[tex]\begin{gathered} R=-10x^2+300x+10000 \\ R=-10(15)^2+300(15)+10000 \\ R=-2250+4500+10000 \\ R=12,250 \end{gathered}[/tex]Hence,
The maximum revenue is = $12,250
