Question 987115
For the area of a rectangle,
{{{A=L*W=60}}}
For the diagonal of a rectangle,
{{{L^2+W^2=8^2}}}
{{{L^2+W^2=64}}}
From the area formula,
{{{L=60/W}}}
{{{L^2=3600/W^2}}}
Substituting,
{{{3600/W^2+W^2=64}}}
{{{3600+W^4=64W^2}}}
Substitute,
{{{u=W^2}}}
{{{3600+u^2=64u}}}
{{{u^2-64u+3600=0}}}
This equation does not have a real solution, only a complex solution.
In other words, there is no rectangle that can have an area of 60 and a diagonal of 8.
Check your problem setup and repost.
The minimum diagonal occurs when {{{L=W=sqrt(60)}}} and is equal to {{{D=sqrt(60)*sqrt(2)}}}