Question 833912
A rectangular garden is 30 ft by 40 ft. Part of the garden is removed in order to install a walkway of uniform width around it. The area of the new garden is one-half the area of the old garden. How wide is the walkway?


Area of the entire garden is {{{A= 30ft * 40ft = 1200ft^2}}}

Let {{{x}}} be the width of the walkway.

Length of the reduced garden is {{{L= (40 - 2x)ft}}}
Width of the reduced garden is {{{ W= (30 - 2x)ft}}}

Area of the reduced garden is {{{A[r] = (40 - 2x) * (30 - 2x)ft^2}}}


Area of the reduced garden is one-half the area of the old garden; so, we have
{{{A[r] =A/2}}} or
{{{2A[r] =A}}}
since
{{{A[r] = (40 - 2x) * (30 - 2x)ft^2}}} and {{{A= 1200ft^2}}}, we have


{{{2(40 - 2x) * (30 - 2x)ft^2=1200ft^2}}}...simplify

{{{(40 - 2x) * (30 - 2x)=600}}}....expand

{{{120 -80x- 60x + 4x^2=600}}}

{{{ 4x^2-140x+1200=600}}}

{{{ 4x^2-140x+1200-600=0}}}

{{{ 4x^2-140x-600=0}}}...both sides divide by {{{4}}}

{{{ x^2-35x-150=0}}}......solve for {{{x}}}


*[invoke quadratic_formula 4, -140, 600, "x"] 


If {{{x = 5}}}, the length is {{{L= (40 - 2*5) =40- 10=30}}} 

and width {{{W= (30 - 2*5) =30-10=20}}}

If {{{x = 30}}}, the length is {{{L= (40 - 2*30) =40- 60=-20}}} 

and width {{{W= (30 - 2*30) = -30}}}

So, the second case ({{{x = 30}}}) is ruled out because you can't have a width and a length equal to negative number.


so the walkway is {{{x=5ft}}} wide