Find two positive numbers $x$ and $y$ such that $x+y=60$ and $x y^{3}$ is maximum.

If $x + y = 16$ and $x^2 + y^2$ is minimum,then the values of $x$ and $y$ are

$A$ wire of length $20 \ m$ is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in $meters$) of the hexagon,so that the combined area of the square and the hexagon is minimum,is:

Let $f:[-1,1] \rightarrow R$ be defined as $f(x)=ax^{2}+bx+c$ for all $x \in[-1,1],$ where $a, b, c \in R$ such that $f(-1)=2, f^{\prime}(-1)=1$ and for $x \in(-1,1)$ the maximum value of $f^{\prime\prime}(x)$ is $\frac{1}{2}.$ If $f(x) \leq \alpha$ for all $x \in[-1,1],$ then the least value of $\alpha$ is equal to:

Let $f(x) = \begin{cases} |x|, & 0 < |x| \le 2 \\ 1, & x = 0 \end{cases}$,then at $x = 0$,$f$ has

If $6x - x^2 + 12$ attains its extreme value $\beta$ at $x = \alpha$, then $\beta =$