i. First, we can use the Revenue formula \(\text{Revenue} = (\text{Price})(\text{Quantity})\) to describe the ticket price:
\(\text{Price} = 20 + 2x\)
\(\text{Quantity} = 1200 - 60x\)
\(\text{Revenue} = (20 + 2x)(1200 - 60x)\)
Therefore, we can determine that the revenue revelation describing the ticket sales is \((20 + 2x)(1200 - 60x)\).
ii. In order to find the maximum revenue (or max point) for the ticket sales, we must complete the square.
We can expand the expression using distributive property, collect like terms and rearrange in descending order:
\(\text{Revenue} = (20)(1200) + (20)(-60x) + (2x)(1200) + (2x)(-60x)\)
\(\text{Revenue} = 24000 - 1200x + 2400x - 120x^2\)
\(\text{Revenue} = -120x^2 + 1200x + 24000\)
Next, as \(a = -120\), we can factor that value out of the first 2 terms:
\(\text{Revenue} = -120(x^2 - 10x) + 24000\)
Then, as \(b = -10\), we can use this value to determine the special number:
\((\cfrac{b}{2})^2\) =
\((\cfrac{-10}{2})^2\) =
\(\cfrac{100}{4}\) =
\(25\)
We can add the special number to the brackets and subtract the expression by the special number multiplied by factor \(a\). We can then simplify the constants:
\(\text{Revenue} = -120(x^2 - 10x + 25) + 24000 - (-120)(25)\)
\(\text{Revenue} = -120(x^2 - 10x + 25) + 24000 + 3000\)
\(\text{Revenue} = -120(x^2 - 10x + 25) + 27000\)
We can factor the trinomial inside the brackets by finding 2 numbers that result in the sum of the second term and the product of the third term:
\(\text{Revenue}= -120(x^2 - 5x - 5x + 25) + 27000\)
\(\text{Revenue}= -120(x(x - 5) -5(x - 5)) + 27000\)
\(\text{Revenue}= -120(x - 5)^2 + 27000\)
As \(x = 5\), we can plug this value into the Price formula to determine what the new ticket prices are:
\(\text{Price} = 20 + 2x = 20 + (2)(5) = 20 + 10 = 30\)
Therefore, we can determine that the selling price to maximize revenue is \($30/\text{ticket}\).
iii. We can also use the x-value to determine the maximum amount of tickets that will be sold using the Quantity formula:
\(\text{Quantity}= 1200 - 60x = 1200 - (60)(5) = 1200 - 300 = 900\)
Therefore, we can determine that we can sell \(900\) tickets at an adjusted price of \($30\).
iv. As \(x = 5\), we can plug this value into the Revenue formula to determine what the new ticket prices are:
\(\text{Revenue} = (20 + (2)(5))(1200 - (60)(5))\)
\(\text{Revenue} = (20 + 10)(1200 - 300)\)
\(\text{Revenue} = (30)(900) = 27000\)
Therefore, we can determine that the max revenue will be \($27000\).