Question 312445
{{{P=2*(L+W)=32}}}
1.{{{L+W=16}}}
The diagonal is then,
2.{{{D^2=L^2+W^2}}}
From the first equation,
{{{L+W=16}}}
{{{L=16-W}}}
Substituting into the second equation,
{{{D^2=(16-W)^2+W^2}}}
{{{D^2=(256-32W+W^2)+W^2}}}
{{{D^2=2W^2-32W+256}}}
Now the distance squared is a function only of W.
Take the derivative wrt W and set it equal to zero to find the minimum.
{{{2*D*(dD/dW)=4W-32=0}}}
Since D cannot be zero (since L and W are both not equal to zero), then {{{dD/dW}}} must be zero when {{{4W-32=0}}}.
{{{4W-32=0}}}
{{{4W=32}}}
{{{W=8}}}
Then from above,
{{{L=16-8=8}}}
The minimum diagonal is formed by an 8m x 8m square.