Problem 1: Rectangle of max area inscribed in a circle Circle radius r. Rectangle sides 2a, 2b with $a^2$ + $b^2$ = $r^2$. Area = 4ab. By AM-GM: $a^2$ + $b^2$ >= 2ab, so ab <= $r^2$/2. Max area = 2$r^2$ (square).

Problem 2: Open box of max volume from a square sheet Sheet side a. Cut squares of side x from corners. V = x(a-2x)^2. V' = (a-2x)(a-6x) = 0. x = a/6 gives max volume = 2$a^3$/27.

Problem 3: Cylinder of max volume inscribed in a cone Cone height H, radius R. Cylinder radius r, height h. By similar triangles: h = H(1-r/R). V = pi*$r^2$h = piH*$r^2$(1-r/R). V' = 0 gives r = 2R/3. Max V = 4pi*$R^2$*H/27.

Problem 4: Closest point on a curve to a given point Minimize distance squared $d^2$ = (x-a)^2 + (f(x)-b)^2. Set d$\frac{d^2}{dx}$ = 0 and solve.

Problem 5: Maximum area of triangle with vertex on a curve Base on x-axis. Vertex at (x, f(x)). Area = $\frac{1}{2}$baseheight.