Question 337862
<pre>
Let the length be x
Let the width by y

Area = xy = 80
Perimeter = 2x + 2y <u><</u> 40 

So we have 

{{{system(xy=80,2x+2y<=40)}}}

Solve the equation for y {{{y=80/x}}}

Substitute in the inequality

{{{2x + 2(80/x) <=40}}}

{{{2x + 160/x <= 40}}}

Since x is positive we can multiply through by x without
reversing the inequality symbol

{{{2x^2+160 <= 40x}}}

{{{2x^2-40x+160<=0}}}

Divide through by 2

{{{x^2-20x+80<=0}}}

Find the zeros of the expression on the left.

It doesn't factor, so:

 {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

 {{{x = (-(-20) +- sqrt((-20)^2-4*1*80 ))/(2*1) }}}

 {{{x = (20 +- sqrt(400-320))/2 }}}

 {{{x = (20 +- sqrt(80))/2 }}}

{{{x = (20 +- sqrt(16*5))/2 }}}

{{{x = (20 +- 4sqrt(5))/2 }}}

{{{x = (4(5 +- sqrt(5)))/2 }}}

{{{x = 2(5 +- sqrt(5)) }}}

The two zeros are approximately 5.5 and 14.5

So we mark those two points on a number line:


{{{number_line( 700, -1,20, 2(5+sqrt(5)), 2(5-sqrt(5)) )}}}

So we test a point is the region left of 5.5, say 5, in

the region left of the 5 in the inequality:

{{{x^2-20x+80<=0}}}
{{{5^2-20(5)+80<=0}}}
{{{25-100+80<=0}}}
{{{5<=0}}}

That's false so we do not shade the part of the number line
left of the green point.

Next we test a point is the region between 5.5 and 14.5, say 6, in
the inequality:

{{{x^2-20x+80<=0}}}
{{{6^2-20(6)+80<=0}}}
{{{36-120+80<=0}}}
{{{-4<=0}}}

That's true so we shade the part of the number line between the
green and red points:

{{{drawing(700,100,-1,20,-1,1, line(2(5+sqrt(5)), .3, 2(5-sqrt(5)),.3),
line(2(5+sqrt(5)), .25, 2(5-sqrt(5)),.25), line(2(5+sqrt(5)), .2, 2(5-sqrt(5)),.2),number_line( 700, -1,20, 2(5+sqrt(5)), 2(5-sqrt(5)) )  )}}}

Now we test a point is the region right of 14.5, say 15, in

the region left of the 5 in the inequality:

{{{x^2-20x+80<=0}}}
{{{15^2-20(15)+80<=0}}}
{{{225-300+80<=0}}}
{{{5<=0}}}

That's false so we do not shade the part of the number line
right of the red point.

So the solution set for the inequality in interval notation is

{{{solution_set(2(5+sqrt(5)), 2(5-sqrt(5)) )}}}

So yes we can make the length any value  of x between those two
values, and the width {{{y=80/x}}}, and 40 will be enough fencing.

Edwin</pre>