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Given a function \(f\), the difference quotient of \(f\) is the expression
\[ \frac {f(x+h) - f(x)}{h} \]
We will revisit this concept in Chapter IntroductiontoDerivatives, but for now, we use it as a way to practice function notation and function arithmetic. For
reasons which will become clear in Calculus, ‘simplifying’ a difference quotient means rewriting it in a form where the ‘\(h\)’ in the
definition of the difference quotient cancels from the denominator. Once that happens, we consider our work to be
done.
Find and simplify the difference quotients for the following functions
\(f(x) = x^2-x-2\)
\(g(x) = \frac {3}{2x+1}\)
\(r(x) = \sqrt {x}\)
To find \(f(x+h)\), we replace every occurrence of \(x\) in the formula \(f(x) = x^2-x-2\) with the quantity \((x+h)\) to get
In order to cancel the ‘\(h\)’ from the denominator, we rationalize the numerator by multiplying by its conjugate. (Rationalizing the numerator!? How’s that for a twist!)
Since we have removed the original ‘\(h\)’ from the denominator, we are done.
As mentioned before, we will revisit difference quotients in Section LinearFunctions where we will explain them geometrically. For now, we
want to move on to some classic applications of function arithmetic from Economics and for that, we need to think
like an entrepreneur. (Not really, but “entrepreneur” is the buzzword of the day and we’re trying to be
trendy.)
Suppose you are a manufacturer making a certain product. (Poorly designed resin Sasquatch statues, for example. Feel
free to choose your own entrepreneurial fantasy.) Let \(x\) be the production level, that is, the number of items produced in a
given time period. It is customary to let \(C(x)\) denote the function which calculates the total cost of producing the \(x\) items. The
quantity \(C(0)\), which represents the cost of producing no items, is called the fixed cost, and represents the amount of money
required to begin production.
Associated with the total cost \(C(x)\) is cost per item, or average cost, denoted \(\overline {C}(x)\) and read ‘\(C\)-bar’ of \(x\). To compute \(\overline {C}(x)\), we take the total
cost \(C(x)\) and divide by the number of items produced \(x\) to get
\[ \overline {C}(x) = \frac {C(x)}{x}\]
On the retail end, we have the price\(p\) charged per item. To simplify the dialog and computations in this text, we assume that
the number of items sold equals the number of items produced. From a retail perspective, it seems natural to think of the
number of items sold, \(x\), as a function of the price charged, \(p\). After all, the retailer can easily adjust the price to sell more
product.
In the language of functions, \(x\) would be the dependent variable and \(p\) would be the independent variable or,
using function notation, we have a function \(x(p)\). While we adopt this convention elsewhere in the text, (See
Example demandfunctionofprice in Section InverseFunctions.) we will hold with tradition at this point and consider the price \(p\) as a function of the number
of items sold, \(x\). That is, we regard \(x\) as the independent variable and \(p\) as the dependent variable and speak
of the price-demand function, \(p(x)\). Hence, \(p(x)\) returns the price charged per item when \(x\) items are produced and
sold.
Our next function to consider is the revenue function, \(R(x)\). The function \(R(x)\) computes the amount of money collected as a result of
selling \(x\) items. Since \(p(x)\) is the price charged per item, we have \(R(x)= x p(x)\). Finally, the profit function, \(P(x)\) calculates how much money is
earned after the costs are paid. That is, \(P(x) = (R-C)(x) = R(x) - C(x)\). We summarize all of these functions below.
Summary of Common Economic Functions
Suppose \(x\) represents the quantity of items produced and sold.
The price-demand function \(p(x)\) calculates the price per item.
The revenue function \(R(x)\) calculates the total money collected by selling \(x\) items at a price \(p(x)\), \(R(x) = x \, p(x)\).
The cost function \(C(x)\) calculates the cost to produce \(x\) items. The value \(C(0)\) is called the fixed cost or start-up cost.
The average cost function \(\overline {C}(x) = \frac {C(x)}{x}\) calculates the cost per item when making \(x\) items. Here, we necessarily assume \(x > 0\).
The profit function \(P(x)\) calculates the money earned after costs are paid when \(x\) items are produced and sold, \(P(x) = (R-C)(x) = R(x) - C(x)\).
It is high time for an example.
Let \(x\) represent the number of dOpi media players (‘dOpis’ (Pronounced ‘dopeys’ …) ) produced and sold in a typical week.
Suppose the cost, in dollars, to produce \(x\) dOpis is given by \(C(x) = 100x + 2000\), for \(x \geq 0\), and the price, in dollars per dOpi, is given by \(p(x) = 450-15x\) for
\(0 \leq x \leq 30\).
Find and interpret \(C(0)\).
Find and interpret \(\overline {C}(10)\).
Find and interpret \(p(0)\) and \(p(20)\).
Solve \(p(x) = 0\) and interpret the result.
Find and simplify expressions for the revenue function \(R(x)\) and the profit function \(P(x)\).
Find and interpret \(R(0)\) and \(P(0)\).
Solve \(P(x) = 0\) and interpret the result.
We substitute \(x=0\) into the formula for \(C(x)\) and get \(C(0) = 100(0) + 2000 = 2000\). This means to produce \(0\) dOpis, it costs \(\$2000\). In other words, the fixed
(or start-up) costs are \(\$2000\). The reader is encouraged to contemplate what sorts of expenses these might be.
Since \(\overline {C}(x) = \frac {C(x)}{x}\), \(\overline {C}(10) = \frac {C(10)}{10} = \frac {3000}{10} = 300\). This means when \(10\) dOpis are produced, the cost to manufacture them amounts to \(\$ 300\) per dOpi.
Plugging \(x=0\) into the expression for \(p(x)\) gives \(p(0) = 450 - 15(0) = 450\). This means no dOpis are sold if the price is \(\$450\) per dOpi. On the other
hand, \(p(20) = 450-15(20) = 150\) which means to sell \(20\) dOpis in a typical week, the price should be set at \(\$150\) per dOpi.
Setting \(p(x) = 0\) gives \(450-15x = 0\). Solving gives \(x = 30\). This means in order to sell \(30\) dOpis in a typical week, the price needs to be set to
\(\$ 0\). What’s more, this means that even if dOpis were given away for free, the retailer would only be able to move \(30\)
of them. (Imagine that! Giving something away for free and hardly anyone taking advantage of it …)
To find the revenue, we compute \(R(x) = x p(x) = x (450 - 15x) = 450x - 15x^2\). Since the formula for \(p(x)\) is valid only for \(0 \leq x \leq 30\), our formula \(R(x)\) is also restricted to \(0 \leq x \leq 30\).
For the profit, \(P(x) = (R-C)(x) = R(x) - C(x)\). Using the given formula for \(C(x)\) and the derived formula for \(R(x)\), we get \(P(x) = \left (450x - 15x^2\right ) -(100x+2000) = -15x^2+350x-2000\). As before, the validity of this
formula is for \(0 \leq x \leq 30\) only.
We find \(R(0) = 0\) which means if no dOpis are sold, we have no revenue, which makes sense. Turning to profit, \(P(0) = -2000\) since \(P(x) = R(x) - C(x)\)
and \(P(0) = R(0) - C(0) = -2000\). This means that if no dOpis are sold, more money (\(\$2000\) to be exact!) was put into producing the dOpis than
was recouped in sales. In number fixedcostex, we found the fixed costs to be \(\$2000\), so it makes sense that if we sell no dOpis,
we are out those start-up costs.
Setting \(P(x) = 0\) gives \(-15x^2+350x-2000 = 0\). Factoring gives \(-5(x-10)(3x-40) = 0\) so \(x = 10\) or \(x = \frac {40}{3}\). What do these values mean in the context of the problem? Since
\(P(x) = R(x) - C(x)\), solving \(P(x) = 0\) is the same as solving \(R(x) = C(x)\). This means that the solutions to \(P(x) = 0\) are the production (and sales) figures
for which the sales revenue exactly balances the total production costs. These are the so-called ‘break even’
points. The solution \(x=10\) means \(10\) dOpis should be produced (and sold) during the week to recoup the cost of
production. For \(x = \frac {40}{3} = 13.\overline {3}\), things are a bit more complicated. Even though \(x = 13.\overline {3}\) satisfies \(0 \leq x \leq 30\), and hence is in the domain of \(P\), it
doesn’t make sense in the context of this problem to produce a fractional part of a dOpi. (We’ve seen this
sort of thing before in Section modeling.) Evaluating \(P(13) = 15\) and \(P(14) = -40\), we see that producing and selling \(13\) dOpis per week makes
a (slight) profit, whereas producing just one more puts us back into the red. (While breaking even is nice,
we ultimately would like to find what production level (and price) will result in the largest profit. We invite the
reader to revisit the tools in Section QuadraticFunctions and find the answer.)