The lengths of the sides of a triangle are  $10+x^2,10+x^{2}and20-2x^2$. The value of x for which area of triangle is maximum is_____

$$\begin{gathered} s=\frac{10+n^2+10+n^2+20-2n^2}{2}=20 \\ s-a=10-n^2;s-b=10-n^2;s-c=2n^2 \\ \Delta =\sqrt{\left[20(10-n^2)(10-n^2)\left(2n^2\right)\right]} \\ =\sqrt{\left[40{(10-n^2)}^2{n}^2\right]} \\ Let n^2=t \\ Let f\left(t\right)={(10-t)}^{2}t \\ =t^3-20t^2+100t \\ f\text{'}\left(t\right)=3t^2-40t+100=0 \\ \Rightarrow t=10,\frac{10}{3} \\ Max, at t=\frac{10}{3}i.e\mathrm{..}{n}^2=\frac{10}{3}(\therefore f\text{'}\text{'}<0) \end{gathered}$$