# How can functions be used to solve real-world situations?

Oct 26, 2016

I'm going to use a few examples.

The population of Wyoming at the end of 2014 was $584 , 513$. The population increases at an annual rate of 0.2%.

a) Determine an equation for the population with respect to the number of years after $2014$, assuming that the population increases at a constant rate.

Solution

The function will be of the form $p = a {r}^{n}$, where $a$ is the initial population, $r$ is the rate of increase, $n$ is the time in years and $p$ is the population.

$p = 584 , 513 {\left(1.0002\right)}^{n}$

b) Using this model, estimate the population of Wyoming in $17$ years.

solution

Here, $n = 17$.

$p = 584 , 513 {\left(1.002\right)}^{17}$

$p = 604 , 708$

c) Using this model, determine in how many years it will take for the population to exceed $600 , 000$.

Solution

Here, $p = 600 , 000$

$600 , 000 = 584 , 513 {\left(1.002\right)}^{n}$

$1.026495561 = {\left(1.002\right)}^{n}$

$\ln \left(1.026495561\right) = \ln {\left(1.002\right)}^{n}$

$\ln \left(1.026495561\right) = n \ln \left(1.002\right)$

$n = 13$

It will take $13$ years for the population to exceed $600 , 000$.

A vending machine currently sells packages of chips that sell for $2.00 and they sell 100 packages. They find that for each 25 cent increase, they lose five customers. a) Write a function, in standard form, to represent this problem. Make sure that the dependant variable is "Profit". Solution We can use the following formula: $P = \text{number of packages sold" xx "price/package}$$P = \left(100 - 5 x\right) \left(2.00 + 0.25 x\right)$$P = 200 - 10 x + 25 x - \frac{5}{4} {x}^{2}$$P = - \frac{5}{4} {x}^{2} + 15 x + 200$b) Determine the profit after $10$increases in price. Solution Here, $x = 10$. $P = - \frac{5}{4} {\left(10\right)}^{2} + 15 \left(10\right) + 200$$P = - \frac{5}{4} \left(100\right) + 90 + 200$$P = - 125 + 90 + 200$$P = 165$Hence, the profit after $10$increases in price will be $165.

c) Find the optimum price for the chips to maximize the profit for the company.

Solution

This will be the vertex of the function. We will need to complete the square.

$P = - \frac{5}{4} {x}^{2} + 15 x + 200$

$P = - \frac{5}{4} \left({x}^{2} - 12 x\right) + 200$

$P = - \frac{5}{4} \left({x}^{2} - 12 x + 36 - 36\right) + 200$

$P = - \frac{5}{4} {\left(x - 6\right)}^{2} + 45 + 200$

$P = - \frac{5}{4} {\left(x - 6\right)}^{2} + 245$

The optimum price is \$6 per bag.

Hopefully this proves that functions can effectively model real world problems, and can be used to solve them!

Hopefully this helps!