Question 224750
First, look at the entire rectangular vacant 40' by 80'lot.
Its area is:
{{{A[1] = 80*40}}}
{{{A[1] = 3200}}}sq.ft.
Half of this would be:
{{{A[2] = A[1]/2}}}
{{{A[2] = 1600}}}sq.ft.
A diagram would probably help you visualize the problem.
Draw a rectangle and, inside of that one, draw a smaller rectangle a uniform distance on all sides from the larger rectangle.
Now the larger rectangle L = 80' and W = 40" and its area is 3200 sq.ft.
The uniform inside distance between the two rectangles we'll call x, and this will be the number of feet at which Bob will stop mowing to turn the job over to Tom.
The area of the inside rectangle is to be 1600 sq.ft.
So we need to express the area of the inside rectangle in terms of the distance x.
The area of the interior rectangle can be expressed as:
{{{A[2] = (80-2x)(40-2x)}}} and this is to be 1600 sq.ft., so...
{{{(80-2x)(40-2x) = 1600}}} Perform the indicated multiplication on the left side.
{{{3200-240x+4x^2 = 1600}}} Subtract 1600 from both sides and rearrange.
{{{4x^2-240x+1600 = 0}}} Here's your quadratic equation. Factor out a 4 to ease the calculations a bit.
{{{4highlight_green((x^2-60x+400) = 0)}}} Now we need to solve the enclosed quadratic for x.
{{{x^2-60x+400 = 0}}} Since we can't factor this we'll use the quadratic formula to solve:{{{x = (-b+-sqrt(b^2-4ac))/2a}}} where: a = 1, b = -60, and c = 400. Making the appropriate substitutions, we get:
{{{x = (-(-60)+-sqrt((-60)^2-4(1)(400)))/2(1)}}} Evaluate.
{{{x = (60+-sqrt(3600-1600))/2}}}
{{{x = (60+-sqrt(2000))/2}}}
{{{x = (60+20sqrt(5))/2}}} or {{{x = (60-20sqrt(5))/2}}}
{{{cross(x = 52.36)}}} or {{{highlight_green(x = 7.64)}}} Discard the extraneous (larger) solution.
Bob should stop mowing at a uniform distance of 7.64 feet.