We have $$4$$ fences of $$10$$m each and we must limit a garden. How many square meters will the garden optimally have?
Development:
We have $$4$$ fences of $$10$$m and we will have to form of rhombus. The problem can be translated, mathematically, to maximize the area of a rhombus with the fixed sides ($$10$$m).
I construct the function to be maximized. The side of the rhombus is $$10$$m. We call $$y$$ the 'long' diagonal and $$x$$ the 'short' one. $$$A(x,y)=\dfrac{x\cdot y}{2}$$$
I find the relationships. In this case the length of the side $$L$$ is our target, since it must always be $$10$$m. We have to calculate $$L$$ in terms of $$x$$ and $$y$$ so that it is equal to $$10$$.
As $$L=10$$ is the side of the rhombus, we must relate $$L$$ with $$x$$ and $$y$$. Let's use Pythagoras' theorem $$$L^2=\Big(\dfrac{x}{2}\Big)^2+\Big(\dfrac{y}{2}\Big)^2 \rightarrow \dfrac{1}{2}\sqrt{x^2+y^2}=10 \rightarrow y=\sqrt{20^2-x^2}$$$
Rewrite the function $$$A(x,y)=\dfrac{x\cdot y}{2} \rightarrow A(x)=\dfrac{x\cdot \sqrt{20^2-x^2}}{2}$$$ We maximize. To do so, the derivative is equal to zero. $$$A'(x)=\dfrac{\sqrt{20^2-x^2}+x\dfrac{1}{2}(20^2-x^2)^{-1/2}(-2x)}{2}= \dfrac{\sqrt{20^2-x^2}-\dfrac{2x^2}{2\sqrt{20^2-x^2}}}{2}=$$$ $$$=\dfrac{\sqrt{20^2-x^2}}{2}-\dfrac{x^2}{2\sqrt{20^2-x^2}}$$$
Let's do the calculation $$$0=A'=\dfrac{\sqrt{20^2-x^2}}{2}-\dfrac{x^2}{2\sqrt{20^2-x^2}}$$$ $$$(20^2-x^2)-x^2=0 \Rightarrow 2x^2=20^2 \Rightarrow x=10\sqrt{2}$$$
It has been solved and we show that $$$x=10\sqrt{2} \ \mbox{m}$$$
We now find the value of $$y$$ $$$y=\sqrt{20^2-x^2} \rightarrow y=\sqrt{20^2-10^2\cdot 2}=\sqrt{200}=10\sqrt{2}$$$
The rhombus with a maximum area will be the one that has the equal diagonals, in fact, the garden will need to have a square form. In this case, the area will be $$100 \ \mbox{m}^2$$.
Solution:
$$A_{max}=100 \ \mbox{m}^2$$