Optimitzation

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 $100 \ \mbox{m}^2$.