Problem source

A small goat is tethered by a rope to a point at ground level on a side of a square barn  which stands in a large horizontal field of grass. The sides of the barn are of length $2a$ and  the rope is of length $4a$.  Let $A$ be the area of the grass that the goat can graze. Prove that $A\le14\pi a^2$ and determine the minimum value of $A$.

Solution source

\begin{center}
    \begin{tikzpicture}[scale=1]
        \coordinate (A) at (0,0);
        \coordinate (B) at (2,0);
        \coordinate (C) at (2,2);
        \coordinate (D) at (0,2);

        \coordinate (X) at (2,1.2);

        \draw[dashed] (A) -- ({-(4-1.2-2)},0);
        \draw[dashed] (B) -- (2,{-(4-1.2)});
        \draw[dashed] (C) -- (2,{2+4-0.8});
        \draw[dashed] (D) -- ({-(4-0.8-2)},2);


        \draw (A) -- (B) -- (C) -- (D) -- cycle;
        \filldraw (X) circle (1pt);

        \node[right] at ($(X)!0.5!(C)$) {$x$};
        \node[right] at ($(X)!0.5!(B)$) {$2a-x$};

        \node at (4,1.2) {$8\pi a^2$};

        \begin{scope}
            \clip (2,-5) rectangle (6,6);
            \draw (X) circle (4);
        \end{scope}
        \begin{scope}
            \clip (-5,2) rectangle (2,8);
            \draw (C) circle ({4-0.8});
        \end{scope}
        \begin{scope}
            \clip (-2,-5) rectangle (D);
            \draw (D) circle ({4-0.8-2});
        \end{scope}
        \begin{scope}
            \clip (-2,-5) rectangle (B);
            \draw (B) circle ({4-1.2});
        \end{scope}
        \begin{scope}
            \clip (-2,0) rectangle (0,2);
            \draw (A) circle ({4-1.2-2});
        \end{scope}
        
    \end{tikzpicture}
\end{center}

The areas are $8\pi a^2 + \frac14 \pi (4a-x)^2 + \frac14 \pi (2a-x)^2 + \frac14\pi(2a+x)^2+\frac14 \pi x^2$ ie

\begin{align*}
A &= \frac{\pi}{4} \left ( x^2 \left (1 + 1 + 1 + 1 \right) + x \left (4a-4a-8a \right)+\left (32a^2+16a^2+4a^2+4a^2 \right)\right) \\
&= \frac{\pi}{4} \left (4x^2-8ax+56a^2 \right) \\
&= \pi(x^2-2ax+14a^2) \\
&= \pi ((x-a)^2+13a^2)
\end{align*}

Since $x \in [0, 2a]$ we have $13\pi a^2 \leq A \leq 14 \pi a^2$