# Solving Word Problems Involving Quadratic Functions Part Two

**Solving Word Problems Involving Quadratic Functions Part Two**

**This hub is a sequel to the hub “Solving Word Problems Involving Quadratic Functions. In this hub I present three additional application problems. I hope you will enjoy this hub and benefit much from it.**

**Sample Problem Number One :**

**The sum of two numbers is 48. Find the maximum product and the two numbers.**

**Solution :**

**Let n = The first number**

**48 – n = the second number**

**f(n) = product of the two numbers**

**f(n) = n (48 – n ) = 48n – n^2**

**f(n) = 48n - n ^2**

**Find the vertex of the parabola:**

** a = -1 (coefficient of n^2**

** b = 48 (coefficient of n)**

** h = -b/2a =****è this is the abscissa of the vertex of the parabola**

**The parabola opens downward and the vertex is the maximum point since a is a negative value.**

**Find h : h = -48/2(-1) = 24**

**k = f(24) = 48(24) - 24^2 =1152 - 576 = 576 ****èthis is the ordinate of the vertex.**

**Thus the maximum product is 576. The product of the two numbers n and 48-n must be equal to the maximum product 576. That is :**

** n(48 – n) = 576**

** 48n - n^2 = 576**

** n^2 -48n + 576 = 0**

** (n -24) ^2 = 0**

** n = 24 **

**If n = 24 then 48-n = 24. Therefore the two numbers whose sum is 48 and whose product is a maximum are 24 and 24.**

**Sample Problem Number Two :**

**Find the dimension and the maximum area of a rectangle if its perimeter is 36.**

**Solution ;**

**Given :**

**Let l = length**

**Perimeter P = 36**

**P = 2l + 2w**

**36 = 2l + 2w**

**w = ( 36 -2l)/2**

**A = l* w**

**A = ((36-2l )/2) l**

**f(A) = 36l/2 - 2l^2/2**

**f(A) = 18l - l^2**

**a = -1 b = 18**

**h = -b/2a = -18/2(-1) = 9**

**k = f(9) = 18(9) - 9^2 = 162 - 81 = 81 ****è this is the ordinate of the parabola.**

**Vertex is at maximum therefore 81 is the maximum area.**

**Solving for l :**

**( (36 – 2l)/2)l = 81 Multiplying this equation by two**

**(36 – 2l) l = 162**

**36l – 2l^2 = 162**

**2l^2 -36l + 162 = 0 Dividing by two**

**l^2 -18l + 81 = 0**

**(l -9)^2 = 0**

**l = 9**

**w = (36 - 2l)/2 = (36 -18)/2 = 18/2 = 9**

**The dimensions of the rectangle should be length = 9 and width = 9 to be able to get the maximum area.**

**Sample Problem Number Three :**

**The sum of two numbers is 36. Find the numbers whose sum of the squares is a minimum.**

**Solution:**

**Let n = first number**

**36 –n = second number**

**F(S) = n^2 + (36 –n )^2**

**F(S) = n ^2 + 1,296 - 72n + n^2**

**F(S) = 2n^2 - 72n + 1,296 Dividing this equation by two**

**F(S) = n^2 - 36n + 648**

** a = 1 b = -36**

**h = -(-36)/2 = 18**

**k = f(18) = 18^2 – 36(18) + 648**

**k= 324 -648 + 648 = 324 This is the ordinate of the vertex**

**The vertex of the parabola is at minimum since a is a positive value.**

**324 is the minimum sum of the squares**

**324 = n^2 – 36n + 648**

**n^2 -36n + 648 -324 = 0**

**n^2 -36n + 324 = 0**

**(n – 18)^2 = 0**

**n – 18 = 0 ==****è n = 18**

**The first number is 18 and the other number is 18. The two numbers which will give a minimum value for the sum of the squares are 18 and 18.**

**SOURCE:**

**ADVANCED ALGEBRA, TRIGONOMETRY AND STATISTICS**

**By :**

**Orines**

**Esparrago**

** Reyes **

## More by this Author

- 14
Solving Word Problems Involving Quadratic Equation One of the most important topics in Algebra is solving quadratic equations. It has found wide applications in many areas of Mathematics. This hub presents...

- 18
Solving Word Problems Involving Quadratic Function Some of the most important functions in applications are the quadratic functions. The quadratic functions are one of the simplest type of polynomial...

- 3
Solving Work Problems Involving System of Linear Equations In this hub I presented several challenging work problems involving system of linear equations complete with solution. I hope this hub will be useful...