# Solving Word Problems Involving Quadratic Functions Part Two

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

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**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.**

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**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.**

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**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 **