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Section Assignment #5 - Done In Class on Thursday 2/27

You have a small business which sells boxes of greeting cards. Let \(x\) be the demand for boxes of cards and assume that it is inversely proportional to the square of the price \(p\) of a box of cards.
  1. If you charge $16 per box of cards, 100 boxes are sold. Find an expression for the price \(p\) of a box of cards as a function of the demand \(x\).

    2. From the given information we know \(x=\frac{k}{p^2}\), and that when \(p=16,\; x=100\). It follows that \(k=xp^2=100(16^2)=10^216^2=160^2=25600\). Thus, \[ p=p(x)=\sqrt{\frac{k}{x}}=\frac{160}{\sqrt{x}}. \] Your initial investment for the business is $750 and it costs you $5 for each box of cards.
  2. Find an expression for the cost \(C\) of your business as a function of demand.

    3. \[ C(x)=750+5x, \] the initial investment, plus the cost of every box demanded/sold.
  3. Find an expression for the revenue \(R\) of your business as a function of demand.

    4. \[ R(x)=xp=x\frac{160}{\sqrt{x}}=160\sqrt{x}. \]
  4. Find an expression for the profit \(P\) of your business as a function of demand. (Recall, Profit= Revenue- Cost.)

    5. \[ P(x)=160\sqrt{x}-750-5x. \]
  5. According to your answer to part d, determine which values of demand lead to an increase in profits, and which values of demand lead to a decrease in profits.

    6. To determine where the profit is increasing/decreasing we should determine where its derivative is positive/negative. \[ P'(x)=160\frac{1}{2\sqrt{x}}-0-5=\frac{80}{\sqrt{x}}-5. \] \(P'(x)>0\) if \(80>5\sqrt{x}\). That is for \(x<16^2=256\). Since demand cannot be negative, we conclude that profit is increasing for \(x\in [0,256)\).
    \(P'(x)<0\) if \(80<5\sqrt{x}\). So profit is decreasing when \(x>256\).
  6. What values of demand and price will maximize your profit?

    7. Note that from part e we know that \(P'(256)=0\). This means \(P(x)\) has a horizontal tangent at \(x=256\), which corresponds to either a peak or a valley. Since the derivative tells us profits are initially increasing and then begin to decrease once demand passes 256 boxes, the graph of \(P(x)\) must have a peak at \(x=256\). That is, the profit is maximized when the demand is 256 and the prices is \(\frac{160}{\sqrt{256}}=\frac{160}{\sqrt{16^2}}=10\).