Question 1156708
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Let the width be x; then the length is 1250/x.<br>
The total length of fencing is 4 times the width plus two times the length.<br>
{{{f(x) = 4x+2500/x}}}<br>
Find the minimum amount of fencing by taking the derivative and finding when it is zero.<br>
{{{df/dx = 4-2500/x^2}}}<br>
{{{4-2500/x^2 = 0 }}}
{{{4 = 2500/x^2}}}
{{{x^2 = 2500/4 = 625}}}
{{{x = 25}}}<br>
The dimensions of the field that uses the minimum length of fencing are x and 1250/x:
width x = 25
length 1250/x = 1250/25 = 50<br>
The minimum amount of fencing required is 4(25)+2(50) = 200.<br>
{{{graph(400,400,-10,50,-50,450,4x+2500/x,200)}}}<br>