Q.Show that semi-vertical angle of right circular cone of given surface area and maximum volume is $\sin^{-1}\left(\frac{1}{3}\right)$.

The problem is a classic “best seat in a movie theater” optimization. You have a screen that starts 10 feet above the floor and ends 30 feet above the floor (since it’s 20 feet tall). Your eye height is at some fixed level — here we assume you stand on the floor, so your eye is roughly at floor level (or we can treat the floor as the reference). The angle $\theta$ is the angle subtended by the screen at your eye. As you move closer to the wall, the screen appears larger vertically, but you have to look up more steeply; as you move farther away, the vertical angle shrinks. Somewhere in between, the angle is maximized.

The key insight: For a fixed vertical segment, the angle subtended at a point on a horizontal line is maximized when the point’s horizontal distance is the geometric mean of the distances to the bottom and top of the segment. This is a consequence of the law of sines or the tangent subtraction formula.

1. Set up coordinates.

   Place the wall along the $y$-axis, with the floor at $y=0$. The bottom of the screen is at $y=10$ ft, the top at $y=30$ ft. You stand at a point $(x,0)$ on the floor, $x>0$ feet from the wall. The viewing angle $\theta$ is the angle between the lines from your eye to the top and bottom of the screen.

2. Express $\theta$ in terms of $x$.

   Let $\alpha$ be the angle from horizontal to the top of the screen, and $\beta$ the angle to the bottom. Then

The viewing angle is $\theta = \alpha - \beta$. Using the tangent subtraction formula:

3. Maximize $\tan\theta$ (or $\theta$ itself). Since $\theta$ is acute and $\tan$ is increasing on $(0,\pi/2)$, maximizing $\theta$ is equivalent to maximizing $\tan\theta$. So we maximize

Differentiate with respect to $x$:

Set $f'(x)=0$:

We start with the intercept form of a straight line:

where $p = OA > 0$ and $q = OB > 0$ are the x- and y-intercepts. Since the line passes through the fixed point $(a,b)$ with $a,b>0$, we have:

Our goal is to minimize $S = p + q$ subject to this constraint.

1. Express one variable in terms of the other From $\frac{a}{p} + \frac{b}{q} = 1$, solve for $q$:

So $S(p) = p + \frac{bp}{p - a}$, with $p > a$ (since $q>0$).

2. Rewrite $S(p)$ for AM–GM

Write $p = (p - a) + a$:

Simplify:

So

3. Apply AM–GM inequality For positive numbers $x = p - a$ and $y = \frac{ab}{p - a}$, we have: $x + y \ge 2\sqrt{xy} = 2\sqrt{(p - a) \cdot \frac{ab}{p - a}} = 2\sqrt{ab}.$ …

Concept & Intuition

The quadrilateral PQRS is inscribed in the rectangle — each vertex lies on a different side. The area of the quadrilateral is not fixed; it changes as the vertices slide along the sides. The problem asks for the value of $x$ that makes the quadrilateral’s area as small as possible.

A classic trick: instead of minimising the quadrilateral’s area directly, notice that the quadrilateral is what’s left of the rectangle after cutting off four right‑angled triangles at the corners. The rectangle’s area is constant ($10 \times 8 = 80$ cm²), so minimising the quadrilateral’s area is equivalent to maximising the total area of the four corner triangles.

Each corner triangle is right‑angled, with legs given by the distances marked $x$ and the leftover lengths on the sides. This turns the problem into a simple quadratic maximisation.

Step‑by‑step reasoning

1. Label the rectangle and the triangles

   Rectangle ABCD:

   • Top side AB = 10 cm, left side AD = 8 cm.
   • Q on AB, with AQ = $x$ cm → QB = $10 - x$ cm.
   • R on BC, with BR = $x$ cm → RC = $8 - x$ cm.
   • S on CD, with CS = $x$ cm → SD = $10 - x$ cm.
   • P on DA, with DP = $x$ cm → PA = $8 - x$ cm.

   The four corner triangles are:

   • $\triangle AQP$ (top‑left corner): legs AQ = $x$, AP = $8 - x$.
   • $\triangle BQR$ (top‑right corner): legs BQ = $10 - x$, BR = $x$.
   • $\triangle CRS$ (bottom‑right corner): legs CR = $8 - x$, CS = $x$.
   • $\triangle DPS$ (bottom‑left corner): legs DP = $x$, DS = $10 - x$.

2. Write the total area of the four triangles

   Area of a right triangle = $\frac{1}{2} \times \text{leg}_1 \times \text{leg}_2$.

   So:

3. Simplify the expression

4. Relate to quadrilateral area

5. Minimise the quadrilateral area …

The key idea: For any right triangle, the hypotenuse is the diameter of its circumcircle. So the circle’s area depends only on the hypotenuse length. Given a fixed triangle area, we want the smallest possible hypotenuse — that happens when the legs are equal, making the triangle isosceles right-angled.

1. Relate triangle area to legs. Let the legs be $a$ and $b$. The area is

2. Express the circumcircle’s area in terms of the hypotenuse. In a right triangle, the hypotenuse $c$ is the diameter of the circumcircle. So the radius is $R = c/2$, and the circle’s area is

3. Write $c^2$ in terms of $a$ and $b$. By Pythagoras:

We want to minimize $c^2$ given the product $ab = 18/\pi$.

4. Minimize $a^2 + b^2$ for fixed product. By AM–GM or by symmetry, for a fixed product, the sum of squares is smallest when $a = b$.

Then

$c^2 = a^2 + b^2 = 2a^2 = \frac{36}{\pi}.$ …

Concept and Intuition

We have a solid that is a cylinder topped with a hemisphere. Both share the same radius $x$. The cylinder’s height is given as $20 - 4x$, so the total volume is the sum of the cylinder’s volume and the hemisphere’s volume. The volume is expressed as $V = \frac{1}{3}\pi y$, so $y$ is essentially $3V/\pi$. To maximize $y$, we maximize $V$. The key is to treat $x$ as a variable, write $V(x)$, then use calculus (or algebra) to find the maximum, while respecting that the height must be positive (so $x < 5$) and the radius positive.

Step-by-step solution

1. Write the volume of each part.

   • Cylinder volume: $\pi x^2 \cdot \text{height} = \pi x^2 (20 - 4x)$.
   • Hemisphere volume: half of a sphere of radius $x$, so $\frac{1}{2} \cdot \frac{4}{3}\pi x^3 = \frac{2}{3}\pi x^3$.

2. Total volume $V$ as a function of $x$.

3. Relate $V$ to $y$. Given $V = \frac{1}{3}\pi y$, we have

4. Find the maximum of $y(x)$. Differentiate:

Set derivative to zero: $30x(4 - x) = 0$ gives $x = 0$ (minimum, trivial) or $x = 4$.

5. Check constraints. …

The problem asks for the dimensions of the largest rectangle that can be placed inside a right triangle, with one corner fixed at the right angle. The rectangle’s base lies along the horizontal leg $a$, its left side along the vertical leg $b$, and its top-right corner touches the hypotenuse. This is a classic optimization problem: we want to maximize the area $A = x \cdot y$ subject to the constraint that the point $(x, y)$ lies on the hypotenuse.

Why this approach works:

The hypotenuse is a straight line connecting $(0, b)$ to $(a, 0)$. Any point on it satisfies a linear relation between $x$ and $y$. By expressing $y$ in terms of $x$ (or vice versa), the area becomes a quadratic function of one variable. The maximum of a quadratic occurs at its vertex, which we can find by symmetry or calculus. The result is beautifully simple: the rectangle’s dimensions are exactly half the triangle’s legs.

1. Set up the coordinate system and the line of the hypotenuse. Place the right angle at the origin $(0,0)$. Then the legs lie along the axes: the horizontal leg from $(0,0)$ to $(a,0)$, the vertical leg from $(0,0)$ to $(0,b)$. The hypotenuse connects $(a,0)$ to $(0,b)$. Its equation is:

because the intercept form of a line is $x/a + y/b = 1$.

2. Express the rectangle’s dimensions and area. The rectangle has width $x$ (along the base) and height $y$ (along the left side). Its top-right corner $(x, y)$ lies on the hypotenuse, so $x$ and $y$ satisfy the line equation. Solve for $y$:

The area is:

3. Maximize the area. $A(x)$ is a quadratic in $x$ that opens downward (coefficient of $x^2$ is negative). Its maximum occurs at the vertex. For a quadratic $A(x) = -\frac{b}{a}x^2 + bx$, the vertex is at: $x = -\frac{b}{2 \cdot (-\frac{b}{a})} = \frac{a}{2}$ …

We are told the box has a fixed volume, and we want the "most economical proportion" — meaning the dimensions that use the least material (minimum surface area) for that volume. The base is a rectangle where one side is three times the other. Let the shorter side of the base be $x$, so the longer side is $3x$. Let the height be $h$. The volume $V$ is fixed, so:

We want to minimize the total surface area (including the lid, since it's a covered box). The surface area $S$ consists of:

• Top and bottom: each $3x^2$, so total $2 \cdot 3x^2 = 6x^2$
• Four sides: two of size $x \cdot h$ and two of size $3x \cdot h$, so total $2xh + 2(3x)h = 2xh + 6xh = 8xh$

Thus:

Now we use the fixed volume to eliminate $h$:

Substitute into $S$:

We minimize $S$ with respect to $x$. Take the derivative:

Set to zero:

So:

Now find $h$ from the volume relation:

Simplify: $x^2 = \left( \frac{2V}{9} \right)^{2/3}$, so:

Now $9^{2/3} = (9^{1/3})^2 = (3^{2/3})^2 = 3^{4/3}$. So:

But $x = \left( \frac{2V}{9} \right)^{1/3} = V^{1/3} \cdot 2^{1/3} \cdot 3^{-2/3}$. Compare $h$ and $x$:

Concept: maximise the area of a triangle with two given equal sides; the river bank supplies the third side, so no fencing constraint acts on it.

The two fenced sides each have length x, with an included angle theta between them.

Area A(theta) = (1/2) * x * x * sin theta = (1/2) x^2 sin theta. …

Since $\angle B=90^\circ$, side $b$ (opposite $B$) is the hypotenuse. With $A=90^\circ-C$: $a=b\sin A=b\cos C$ and $c=b\sin C$.

Constraint: $a+b=b\cos C+b=b(1+\cos C)=4\Rightarrow b=\dfrac{4}{1+\cos C}$.

Area $=\tfrac12 ac=\tfrac12 b^2\sin C\cos C=\tfrac14 b^2\sin 2C=\dfrac{4\sin 2C}{(1+\cos C)^2}.$ …

The key idea here is to find where a function reaches its highest point — that’s a classic optimisation problem. For a differentiable function on an open interval like $x > 0$, the maximum (if it exists) occurs at a critical point where the derivative is zero, provided the function changes from increasing to decreasing there.

Why does this particular function matter? $\frac{\log x}{x}$ appears often in comparisons of growth rates — it tells us that $x^{1/x}$ is maximised at $x = e$, a neat fact. But let’s not jump ahead; we’ll find the maximum step by step.

1. Define the function and its domain.

   Let $f(x) = \frac{\log x}{x}$, with $x > 0$. We want the maximum value of $f(x)$.

2. Differentiate $f(x)$.

   Use the quotient rule:

3. Find critical points. Set $f'(x) = 0$:

So $x = e$ is the only critical point in $x > 0$.

4. Check if it’s a maximum. Look at the sign of $f'(x)$ around $x = e$:
   • For $0 < x < e$, $\log x < 1$, so $1 - \log x > 0$, hence $f'(x) > 0$ — function is increasing.
   • For $x > e$, $\log x > 1$, so $1 - \log x < 0$, hence $f'(x) < 0$ — function is decreasing. …