A running track of 440 m is to be laid out enclosing a football field. The football field is in the…

Let the rectangle have length $x$ and breadth $2r$, with a semi-circle of radius $r$ at each end. Perimeter of track: $2x + 2\pi r = 440 \Rightarrow x = 220 - \pi r$ Area of rectangle: $A = x \cdot 2r = 2r(220 - \pi r) = 440r - 2\pi r^2$ $\frac{dA}{dr} = 440 - 4\pi r = 0 \Rightarrow r = \frac{110}{\pi}$ $\frac{d^2A}{dr^2} = -4\pi < 0$, so $A$ is maximum. $x = 220 - \pi\cdot\frac{110}{\pi} = 110\text{ m}$; breadth $2r = \frac{220}{\pi} = 70\text{ m}$ (taking $\pi = \frac{22}{7}$) Area of field = rectangle + two semi-circles $= 110 \times 70 + \pi r^2 = 7700 + \frac{22}{7}\times\left(\frac{110 \times 7}{22}\right)^2 = 7700 + 3850 = 11550\text{ m}^2$