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

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

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Answer 1 · 2016-10-26T21:48:49

I'm going to use a few examples.

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

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

Solution

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

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

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

solution

Here, $n = 17$ $n = 17$ .

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

$p = 604, 708$ $p = 604,708$

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

Solution

Here, $p = 600, 000$ $p = 600,000$

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

$1.026495561 = {(1.002)}^{n}$ $1.026495561 = (1.002)^n$

$ln (1.026495561) = {ln (1.002)}^{n}$ $ln(1.026495561) = ln(1.002)^n$

$ln (1.026495561) = n ln (1.002)$ $ln(1.026495561) = nln(1.002)$

$n = 13$ $n = 13$

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

**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:

$P = number of packages sold \times price/package$ $P = "number of packages sold" xx "price/package"$

$P = (100 - 5 x) (2.00 + 0.25 x)$ $P = (100 - 5x)(2.00 + 0.25x)$

$P = 200 - 10 x + 25 x - \frac{5}{4} x^{2}$ $P = 200 - 10x + 25x - 5/4x^2$

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

b) Determine the profit after $10$ $10$ increases in price.

Solution

Here, $x = 10$ $x= 10$ .

$P = - \frac{5}{4} {(10)}^{2} + 15 (10) + 200$ $P = -5/4(10)^2 + 15(10) + 200$

$P = - \frac{5}{4} (100) + 90 + 200$ $P = -5/4(100) + 90 + 200$

$P = - 125 + 90 + 200$ $P = -125 + 90 + 200$

$P = 165$ $P = 165$

Hence, the profit after $10$ $10$ increases in price will be $$ 165$ $$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.

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

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

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

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

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

The optimum price is $$ 6$ $$6$ per bag.

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

Hopefully this helps!

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

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