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[isabelle2hot]

I previosuly posted this problem but my instructor changed a number. This is what I have so far.

You manage a 220 unit motel. All units will be occupied, on average, when you charge $50 per day per unit. Experience shows that for each $1 per day per unit that you increase the price, you will accrue 2 vacancies. Each occupied room costs $4 per day to clean and supply. On this basis, what price, in dollars, should you charge per unit per day to maximize your profit?

Our revenue (income) is: $220 x $50= $11,000 per day.
. . Each room costs $4 to clean: $ x $220 cost = $880
Our profit is: . $11,000 - $880 = !10,120 per day

Then
P = (50 + x)(220 - 2x) - 4 (220-2x)
P = 10,120 + ?? - 2x^2

This is where I am confused.......

 

[stapel]

You've done the hard part -- the set-up. Now do the easy part: Multiply out the factors, simplify the quadratic, and find the vertex of the parabola to find the maximum point of the profit function. (The "h" part will be the number of cost increases; the "k" part will be the profit.)

Eliz.

 

[soroban]

Hello, isabelle2hot!

Then: \(\displaystyle \;P\:=\50\,+\,x)(220\,-\,2x)\,-\,4(220\,-\,2x)\) . . . . correct!

\(\displaystyle P\:=\:10,120\,+\,??\,-\,2x^2\)

This is where I am confused. . Really?

I don't understand your difficulty . . . it's just Algebra I, isn't it?

\(\displaystyle P\;=\;(50\,+\,x)(220\,-\,2x)\,-\,4(220\,-\,2x)\)

. . . \(\displaystyle =\:11,000\,-\,100x\,+\,220x\,-\,2x^2\,-\,880\,+\,8x\)

. . . \(\displaystyle =\:10,120\,+\,128x\,-\,2x^2\)