1. ## Problem solving with Quadratics x2

Find the width of a uniform concrete path placed around a 30 m by 40 m rectangular lawn given that the concrete has area one quarter of the lawn.

2. Hello,
Say the area of the lawn is $A$, and the area of the path is $A'$. We are given the dimensions of the lawn, so $A = 30 \times 40 = 1200$.

Now you know that $A' = \frac{A}{4} = 300$.

Let us define a function that gives the area of the concrete path with respect to some width, denoted $x$. After investigation, we find that this function is defined by : $f(x) = 60x + 2(40 - 2x)x = 60x + (80 - 4x)x = 60x + 80x - 4x^2 = -4x^2 + 140x$. You are trying to find $x$ such as $f(x) = 300$. Solve $-4x^2 + 140x = 300$ for $x$ :

$-4x^2 + 140x - 300 = 0$

$\Delta = 140^2 - 4 \times (-4) \times (-300) = 14800$

$x_1 = \frac{-140 - \sqrt{14800}}{-8}$

$x_2 = \frac{-140 + \sqrt{14800}}{-8}$

A width can only be positive, so discard the negative solution. You just found the width of the concrete path

Does that make sense ?

3. Originally Posted by Bacterius
Hello,
Say the area of the lawn is $A$, and the area of the path is $A'$. We are given the dimensions of the lawn, so $A = 30 \times 40 = 1200$.

Now you know that $A' = \frac{A}{4} = 300$.

Let us define a function that gives the area of the concrete path with respect to some width, denoted $x$. After investigation, we find that this function is defined by : $f(x) = 60x + 2(40 - 2x)x = 60x + (80 - 4x)x = 60x + 80x - 4x^2 = -4x^2 + 140x$. You are trying to find $x$ such as $f(x) = 300$. Solve $-4x^2 + 140x = 300$ for $x$ :

$-4x^2 + 140x - 300 = 0$

$\Delta = 140^2 - 4 \times (-4) \times (-300) = 14800$

$x_1 = \frac{-140 - \sqrt{14800}}{-8}$

$x_2 = \frac{-140 + \sqrt{14800}}{-8}$

A width can only be positive, so discard the negative solution. You just found the width of the concrete path

Does that make sense ?
Thank you for helping.

I tried the answer you gave me but it doesn't match up with the answer in the back of my math book. The answer i have in my math text book is: 2.026 m.