Using quadratics to find the dimensions of a rectangle

Question

A gardener has 40m of fencing to make three equal rectangular plots. The total area of the three plots is 30m². Determine the constraints on the dimensions of the rectangles, and find the dimensions of each rectangle. 

the answer is 2.28m by 4.39m. could you show the method on how to solve this problem

Thank you. 

Answer 1

If we label the dimensions of each rectangle as x and y, then each rectangle has an area of xy and a perimeter of 2x + 2y.

The total area of the three rectangles is 30, so 3xy = 30

 

The total fencing for the three rectangles gives our other equation 3(2x + 2y) = 40

 

Now we have a system of equations that we can solve.

 

(1) 3xy = 30

(2) 3(2x + 2y) = 40

 

Simplifying each equation gives

 

(1) xy = 10

(2) 6x + 6y = 40

 

Solving equation (2) for y gives

y = 20/3 - x

 

Substitute this expression for y into equation (1) to get

x(20/3 - x) = 10

 

Put this quadratic in standard form to get

x² - (20/3)x + 10 = 0

 

Use the quadratic formula with a = 1, b = -20/3 and c = 10 to get your answers

Answer 2

Here are the two main equations:

 

  3 * perimeter = 40

 

  3 * area = 30

 

---------------------------------

3 * perimeter = 40

3 * (2L + 2w) = 40

 

 3*2*(L+w) = 40

 3*(L+w) = 20

 

L+w = 20/3

 

w = 20/3 - L <--- this is the first equation

 

---------------------------

 

3*Area = 30

 

Area = 30/3 = 10

 

L * w = 10 <---- this is the second equation

 

-----------------------------------

 

L * (20/3 - L) = 10 <--- substitutes first equation into the second equation

 

L * (3L - 20) = -30  <---- multiplies everything by -3 to clear away the fraction

                                     and changes the signs

 

3L^2 - 20L + 30 = 0  <---- adds 30 to both sides

 

L = [20 +or- sqrt(  (-20)^2 - 4(3)(30 ) ] / (2*3)  <--- quadratic formula with A=3, B= -20, C = 30

 

L = [ 20 +or- sqrt( 400 - 360) ]/ 6

 

L = [ 20 +or- sqrt( 40) ]/6

 

L  = [ 20 +or- sqrt(4*10)]/ 6 

 

L = [ 20 +or- sqrt(4)*sqrt(10)]/6

 

L = [ 20 +or- 2*sqrt(10)]/6

 

L = [ 20 +or- 2 * sqrt(10)]/6

 

L = [ 10 +or- sqrt(10)]/3

 

For the positive branch of the solution:

which is approximately 4.38742588672.... or the latter number you have given as the answer.

The width is the area/length = 10 / [( 10 + sqrt(10)/3]

                                          = 30 / [ 10 + sqrt(10)]

 

                                         = 30 ( 10 - sqrt(10)) / [ 10 + sqrt(10][10 - sqrt(10])  <--- rationalizes the denominator

 

                                          = 30 ( 10 - sqrt(10)) /  [ 100 - 10 ]

 

                                         = 30 ( 10 - sqrt(10))/ 90

 

                                         = ( 10 - sqrt(10) ) / 3

 

                                          which is approximately 2.279241 or the former number you have listed as the answer

 

 

 For the negative branch:

  L = [ 10 - sqrt(10)]/3  which is the former number you have given as the answer, as just shown in hte above line.

 

 The width in this case is 10/ [( 10 - sqrt(10))/3] =

                                     30 / [ 10 - sqrt(10)] =

 

                                    30 [ 10 + sqrt(10)] /   [ 10 - sqrt(10)][10 + sqrt(10)]

 

                                   30 [ 10 + sqrt(10)] / [ 100 - 10]

 

                                  30 [ 10 + sqrt(10)] / 90

 

                                      [ 10 + sqrt(10) ]/3 which was the original value for length as shown above in bold, the

                                                                   latter value you have listed as the answer.