Question

A farmer wants to construct a circular garden and a square garden in his field. He wants to keep the sum of their perimeters 600m. Prove that the sum of their areas is the least, when the side of the square garden is double the radius of the circular garden.

Answer 1

Let the radius of circular garden be r (in metres) and the side of the square garden be x (in metres). Then, $600=2\pi r+4x\Rightarrow x=\frac{600-2\pi r}{4}$... (i)
The sum of the areas, $S=\pi r^2+x^2\Rightarrow S=\pi r^2+{\left(\frac{600-2\pi r}{4}\right)}^2$
$\Rightarrow \frac{dS}{dr}=2\pi r-\frac{4\pi }{16}(600-2\pi r)=\frac{\pi }{2}(4r-300+\pi r)$
For local points of maxima and/or minima, $\frac{dS}{dr}=\frac{\pi }{2}(4r-300+\pi r)=0$
$\Rightarrow \pi r=300-4r\Rightarrow r=\left(\frac{300}{\pi +4}\right)$metres.
Also $\frac{d^{2}S}{d{r}^2}=\frac{\pi }{2}(4+\pi )$
$\because {\left[\frac{d^{2}S}{d{r}^2}\right]}_{atr=\frac{300}{\pi +4}}=\frac{\pi }{4}(4+\pi )>0$
$\therefore$ S is least.
By (i), 600 = 2 $\pi r+4x\Rightarrow 600=600-8r+4x\therefore x=2r$
Therefore, when the side of the square garden is double the radius of the circular garden, the sum of their areas is least.
To achieve any goal, there is every possibility that energy, time and money are required to be invested. One must plan in such a manner that least energy, time and money are spent. A good planning and execution, therefore, is essentially needed.