Problems from Optimitzation

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. $$A(x,y)={\displaystyle \frac{x\cdot y}{2}}$$

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 $$L^2=({\displaystyle \frac{x}{2}}{)}^2+({\displaystyle \frac{y}{2}}{)}^2\to {\displaystyle \frac{1}{2}}\sqrt{x^2+y^2}=10\to y=\sqrt{{20}^2-x^2}$$

Rewrite the function $$A(x,y)={\displaystyle \frac{x\cdot y}{2}}\to A(x)={\displaystyle \frac{x\cdot \sqrt{{20}^2-x^2}}{2}}$$ We maximize. To do so, the derivative is equal to zero. $$A^{\prime }(x)={\displaystyle \frac{\sqrt{{20}^2-x^2}+x{\displaystyle \frac{1}{2}}({20}^2-x^2{)}^{-1/2}(-2x)}{2}}={\displaystyle \frac{\sqrt{{20}^2-x^2}-{\displaystyle \frac{2x^2}{2\sqrt{{20}^2-x^2}}}}{2}}=$$ $$={\displaystyle \frac{\sqrt{{20}^2-x^2}}{2}}-{\displaystyle \frac{x^2}{2\sqrt{{20}^2-x^2}}}$$

Let's do the calculation $$0=A^{\prime }={\displaystyle \frac{\sqrt{{20}^2-x^2}}{2}}-{\displaystyle \frac{x^2}{2\sqrt{{20}^2-x^2}}}$$ $$({20}^2-x^2)-x^2=0\Rightarrow 2x^2={20}^2\Rightarrow x=10\sqrt{2}$$

It has been solved and we show that $$x=10\sqrt{2}\text{m}$$

We now find the value of $y$ $$y=\sqrt{{20}^2-x^2}\to y=\sqrt{{20}^2-{10}^2\cdot 2}=\sqrt{200}=10\sqrt{2}$$

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

Solution:

$A_{max}=100{\text{m}}^2$

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