Question 708791
A semester-1 Calculus problem.


x, length in one direction
y, length in other direction

h, length of diagonal


{{{h=(x^2+y^2)^(1/2)}}}
perimeter is 2(x+y)
perimeter is a constant, 10 feet:  {{{2(x+y)=10}}}
IDEA: Use perimeter equation and solve it for y.  Substitute this into the diagonal equation.



{{{(x+y)=10/2}}}
{{{y=5-x}}}


{{{h=(x^2+y^2)^(1/2)}}}
{{{h=(x^2+(5-x)^2)^(1/2)}}}
={{{(x^2+(25-10x+x^2))^(1/2)}}}
{{{h=(2x^2-10x+25)^(1/2)}}}
This last formula or function, you want to see if you can find the minimum, if possible.  Find the derivative and see about setting it equal to zero and this might help. (The function does in fact have a minimum.)