A trapezium has its three sides other than the base each equal to $10$ cm. Its area is maximum when the slant foot length $x$ is

1. Area $A(x)=(x+10)\sqrt{100-x^2}$. 2. $A'(x)=\dfrac{-2x^2-10x+100}{\sqrt{100-x^2}}$. 3. $A'(x)=0 \Rightarrow 2x^2+10x-100=0 \Rightarrow x=5$ or $x=-10$. 4. Distance cannot be negative, so $x=5$. 5. $A''(5)<0$ confirms a maximum. _Source: syllabus Mathematics Ch 6 "Application of Derivatives", p.22_