Q.A window is in the form of a rectangle surmounted by a semi-circular opening. The perimeter of the window is 30 m. Find the dimensions of the window so that it can admit maximum light through the whole opening. OR Prove that the volume of the largest cone that can be inscribed in a sphere is 8/27 of the volume of the sphere.

Substitute the optimal square side $x=\dfrac{2}{3}$ m (found by maximising the volume function) into the length, breadth and height expressions.

From the case study, cutting a square of side $x$ from each corner of the $3\text{ m}\times 8\text{ m}$ sheet and folding up the sides gives a box of:

• Length $= (8-2x)$ m
• Breadth $= (3-2x)$ m
• Height $= x$ m

Maximising $V(x)=x(3-2x)(8-2x)=4x^3-22x^2+24x$ using $V'(x)=12x^2-44x+24=0$ (i.e. $3x^2-11x+6=0$) gives roots $x=3$ or $x=\dfrac{2}{3}$. Since $0<x<1.5$ is required for the box to be valid, the admissible root is $x=\dfrac{2}{3}$, and $V''\left(\tfrac{2}{3}\right)=-28<0$ confirms this is the maximum.

Substituting $x=\dfrac{2}{3}$:

$\text{Length} = 8 - 2\left(\tfrac{2}{3}\right) = 8 - \tfrac{4}{3} = \tfrac{20}{3}\ \text{m}$ …

The optimal square side is the critical point of the volume function that lies in the valid domain and satisfies the second-derivative maximum test.

For a square of side $x$ removed from each corner of the $3\text{ m}\times 8\text{ m}$ sheet, the box volume is:

$V(x) = x(3-2x)(8-2x) = 4x^3 - 22x^2 + 24x, \qquad 0 < x < 1.5$

Differentiating and setting $V'(x)=0$:

$V'(x) = 12x^2 - 44x + 24 = 0 \implies 3x^2 - 11x + 6 = 0$

$x = \frac{11 \pm \sqrt{121-72}}{6} = \frac{11\pm 7}{6} \implies x = 3 \text{ or } x = \frac{2}{3}$

Since the breadth $(3-2x)$ must stay positive, only $x<1.5$ is valid, so $x=3$ is rejected and $x=\dfrac{2}{3}$ is the only admissible critical point.

Confirming it is a maximum: …

The rectangle's top corners lie on the semicircle of radius 30, so the Pythagorean relation between half the length and the breadth gives the breadth as a function of $x$.

Place the centre $O$ of the semicircle at the origin, with the diameter along the $x$-axis. Since the rectangle $PQRS$ is symmetric about $O$ with top side $PQ=x$, the top corners $P,Q$ are at horizontal distance $x/2$ from $O$. Let the breadth (height of the rectangle) be $b$. …

Area = length × breadth, using the breadth found in terms of $x$.

The rectangle has length $PQ=x$ and breadth $b=\sqrt{900-x^2/4}$ (from the semicircle constraint). So the area is …

Maximize $A(x)^2$ (equivalent and algebraically simpler) by setting its derivative to zero.

From $A(x)=x\sqrt{900-x^2/4}$, consider $A^2 = x^2\left(900-\dfrac{x^2}{4}\right) = 900x^2-\dfrac{x^4}{4}$ (maximizing $A^2$ maximizes $A$ since $A\ge0$).

$\frac{d(A^2)}{dx} = 1800x - x^3 = x(1800-x^2)$

Setting this to zero: $x=0$ (rejected, gives zero area) or $x^2=1800 \Rightarrow x=\sqrt{1800}=30\sqrt2$.

…

Subtract the maximum rectangle area from the total semicircular field area.

Total area of the semicircular field: $\dfrac12\pi r^2 = \dfrac12\pi(30)^2 = 450\pi$ sq. m.

At maximum, $x=30\sqrt2$, so breadth $b=\sqrt{900-\dfrac{(30\sqrt2)^2}{4}} = \sqrt{900-\dfrac{1800}{4}} = \sqrt{900-450}=\sqrt{450}=15\sqrt2$.

…

$r^2 = R^2 - x^2$.

From the figure, $O$ is the sphere's centre, $C$ is the centre of the cone's circular base, $OC=x$, $CA=r$ (radius of the cone's base), and $OA=R$ (a radius of the sphere, since $A$ lies on the sphere).

…

$V = \dfrac{\pi}{3}(R+x)^2(R-x)$.

From the figure, the cone's height is $h=R+x$ (from the base at $C$ up to the apex $D$ at the top of the sphere), and from the previous part, $r^2=R^2-x^2$.

Volume of a cone: $V=\dfrac13\pi r^2h$. …