Optimise corner flower crates around a central pond

The horizontal and vertical crate sides are $13-11\sin\theta$ and $13-11\cos\theta$, so

$X=(13-11\sin\theta)(13-11\cos\theta).$

The product rule gives $X^{\prime}=-143\cos\theta+121\cos^2\theta+143\sin\theta-121\sin^2\theta$. Using $\cos^2\theta-\sin^2\theta=(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)$,

$X^{\prime}=11(\sin\theta-\cos\theta)(13-11\sin\theta-11\cos\theta).$

The first factor gives $\tan\theta_1=1$, hence $\theta_1=\pi/4$. The other gives $\sin\theta+\cos\theta=13/11$, or

$\sin(\theta_2+\pi/4)=\frac{13}{11\sqrt2}.$

Thus $\alpha=\pi/4$, $k=13/(11\sqrt2)$ and $\theta_2=\arcsin(13/(11\sqrt2))-\pi/4\approx0.20395638$ rad (printed $0.20396$). The supplementary sine solution is outside $[0,\pi/4]$ and is rejected.

The printed sign charts give a negative-to-positive change at $\theta_2$ and positive-to-negative across $\pi/4$ in the smooth extension. In the actual allowed interval, $X^{\prime}<0$ on $(0,\theta_2)$ and $X^{\prime}>0$ on $(\theta_2,\pi/4)$. Thus $\theta_2$ is a minimum and $\pi/4$ is the right-endpoint maximum candidate. The source’s test immediately to the right of $\pi/4$ lies outside the stated geometric interval; comparing endpoint values below establishes the global maximum without that extra sample.

The printed second derivative is $X^{\prime\prime}=143(\sin\theta+\cos\theta)-484\sin\theta\cos\theta$. At $\pi/4$ this is $143\sqrt2-242\approx-39.8<0$. At $\theta_2$, $\sin\theta+\cos\theta=13/11$ and $\sin\theta\cos\theta=24/121$, so $X^{\prime\prime}=169-96=73>0$ (the source’s rounded-angle calculation prints $72.999$).

The requested area values are $X_{\min}=24\text{ m}^2$ at $\theta_2$, and $X_{\max}=(13-11/\sqrt2)^2=459/2-143\sqrt2\approx27.26746058\text{ m}^2$ at $\pi/4$. The source omits the minimum area despite asking for it, so it is supplied explicitly here. Since $X(0)=26<X(\pi/4)$, this endpoint is also the global maximum on the permitted interval.

The four corner crates do not overlap and their interiors lie outside the pond. Use the maximum crate area, retaining full precision:

$\begin{aligned}&G_{\min}\\ & =26^2-4(13-11/\sqrt2)^2\\ &{}-\pi(11)^2\\ & =572\sqrt2-242\\ &{}-121\pi\approx186.79744659\text{ m}^2.\end{aligned}$

Hence the smallest grass area is $187\text{ m}^2$ to three significant figures, agreeing with the printed answer. The source substitutes the rounded crate area $27.3$; the exact calculation avoids that intermediate rounding.