SOLUTION: A rectangular field has dimesions 50 m by 40 m. A flowerbed is prepared inside the field with a gravel footpath of equal width around it. The cost of preparing the flowerbed and gr

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: A rectangular field has dimesions 50 m by 40 m. A flowerbed is prepared inside the field with a gravel footpath of equal width around it. The cost of preparing the flowerbed and gr      Log On


   



Question 883351: A rectangular field has dimesions 50 m by 40 m. A flowerbed is prepared inside the field with a gravel footpath of equal width around it. The cost of preparing the flowerbed and gravel footpath is Rs.30 and Rs. 20 per sqm respectively amounting to Rs. 52000. Find the width of the footpath.
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
A rectangular field has dimesions 50 m by 40 m.
Area of filed = 50*40 2000 m^2
Let width of footpath be x
Length of flower bed = (50-2x)
breadth of flower bed = (40-2x)
Area of flower bed = (50-2x)(40-2x)
cost of making flower bed = 30/m^2
Area of footpath = area of field - area of flower bed
Area of foot path = (2000-(50-2x)(40-2x))
cost of making foot path = 20/m^2
Total cost = 52000
30(50-2x)(40-2x) +20*(2000-(50-2x)(40-2x))=52000
30(2000-100x-80x+4x^2)+20(2000-(2000-100x-80x+4x^2))=52000
60000-5400x+120x^2+ 20(2000-2000+180x-4x^2)=52000
60000-5400x+120x^2+ 40000-40000+3600x-80x^2)=52000
60000-1800x+40x^2=52000
40x^2-1800x+8000=0
/10
4x^2-180x+800=0
4x^2-160x-20x+800=0
4x(x-40)-20(x-40)=0
(x-40)(4x-20)=0
x=40 OR x=5
it cannot be 40 m since width is 40 m
so width = 5 m