Q.(Continuing the aluminium-box case study of Q.38) What will be the dimensions of the largest box?

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: $V=\dfrac13\pi r^2h$. …