I'm going to use a few examples.
The population of Wyoming at the end of 2014 was 584,513584,513. 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 20142014, assuming that the population increases at a constant rate.
Solution
The function will be of the form p = ar^np=arn, where aa is the initial population, rr is the rate of increase, nn is the time in years and pp is the population.
p = 584,513(1.0002)^np=584,513(1.0002)n
b) Using this model, estimate the population of Wyoming in 1717 years.
solution
Here, n = 17n=17.
p = 584,513(1.002)^17p=584,513(1.002)17
p = 604,708p=604,708
c) Using this model, determine in how many years it will take for the population to exceed 600,000600,000.
Solution
Here, p = 600,000p=600,000
600,000 = 584,513(1.002)^n600,000=584,513(1.002)n
1.026495561 = (1.002)^n1.026495561=(1.002)n
ln(1.026495561) = ln(1.002)^nln(1.026495561)=ln(1.002)n
ln(1.026495561) = nln(1.002)ln(1.026495561)=nln(1.002)
n = 13n=13
It will take 1313 years for the population to exceed 600,000600,000.
**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" xx "price/package"P=number of packages sold×price/package
P = (100 - 5x)(2.00 + 0.25x)P=(100−5x)(2.00+0.25x)
P = 200 - 10x + 25x - 5/4x^2P=200−10x+25x−54x2
P = -5/4x^2 + 15x + 200P=−54x2+15x+200
b) Determine the profit after 1010 increases in price.
Solution
Here, x= 10x=10.
P = -5/4(10)^2 + 15(10) + 200P=−54(10)2+15(10)+200
P = -5/4(100) + 90 + 200P=−54(100)+90+200
P = -125 + 90 + 200P=−125+90+200
P = 165P=165
Hence, the profit after 1010 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 = -5/4x^2 + 15x + 200P=−54x2+15x+200
P = -5/4(x^2 - 12x) + 200P=−54(x2−12x)+200
P = -5/4(x^2 - 12x + 36- 36) + 200P=−54(x2−12x+36−36)+200
P = -5/4(x - 6)^2 + 45 + 200P=−54(x−6)2+45+200
P = -5/4(x - 6)^2 + 245P=−54(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!
