Comprehensive solutions for the Class 12 Maths (NCERT) Miscellaneous Exercise on Chapter 6. Covers complex optimization problems (geometric shapes, cost minimization), advanced rates of change, maxima/minima of transcendental functions, and proofs of monotonicity using the Mean Value Theorem.

Rbse Solutions Class 12 Maths Chapter 6 Miscellaneous

This exercise covers a mix of topics including Maxima/Minima, Rates of Change, and Proving Monotonicity.

1. Maxima and Minima (Functions)

1. Show that $f(x) = \frac{\log x}{x}$ has maximum at $x = e$.

Find the derivative (Quotient Rule):$$f'(x) = \frac{x \cdot \frac{1}{x} – \log x \cdot 1}{x^2} = \frac{1 – \log x}{x^2}$$
Find critical point ($f'(x) = 0$):$$1 – \log x = 0 \implies \log x = 1 \implies x = e$$
Use Second Derivative Test:$$f”(x) = \frac{x^2 \cdot (-\frac{1}{x}) – (1 – \log x) \cdot 2x}{(x^2)^2} = \frac{-x – 2x + 2x \log x}{x^4} = \frac{-3 + 2 \log x}{x^3}$$
Evaluate $f”(e)$:$$f”(e) = \frac{-3 + 2 \log e}{e^3} = \frac{-3 + 2(1)}{e^3} = -\frac{1}{e^3}$$
Since $f”(e) < 0$, the function has a local maximum at $x=e$.

2. Rates of Change

2. Rate of change of area of an isosceles triangle

Let the equal sides be $a$, and the fixed base be $b$. The height $h$ bisects the base.

$$h = \sqrt{a^2 – (b/2)^2} = \frac{1}{2}\sqrt{4a^2 – b^2}$$

Area $A = \frac{1}{2} \cdot b \cdot h = \frac{b}{4}\sqrt{4a^2 – b^2}$.

Given: $\frac{da}{dt} = -3 \text{ cm/s}$. Find $\frac{dA}{dt}$ when $a = b$.

Differentiate $A$ w.r.t. $t$:$$\frac{dA}{dt} = \frac{b}{4} \cdot \frac{1}{2\sqrt{4a^2 – b^2}} \cdot (8a \frac{da}{dt})$$$$\frac{dA}{dt} = \frac{ba}{2\sqrt{4a^2 – b^2}} \frac{da}{dt}$$
Substitute $a = b$ and $\frac{da}{dt} = -3$:$$\frac{dA}{dt} = \frac{b(b)}{2\sqrt{4b^2 – b^2}} (-3) = \frac{b^2}{2\sqrt{3b^2}} (-3)$$$$\frac{dA}{dt} = \frac{b^2}{2\sqrt{3}b} (-3) = \frac{b}{2\sqrt{3}} (-3) = -\frac{3b}{2\sqrt{3}} = -\frac{\sqrt{3}b}{2}$$The area is decreasing at the rate of $\mathbf{\frac{\sqrt{3}b}{2} \text{ cm}^2/\text{s}}$.

3. Intervals of Monotonicity (Trigonometric Function)

3. Intervals for $f(x) = \frac{4 \sin x – 2x – x \cos x}{2 + \cos x}$

This is the same function as in Exercise 6.2, Q9, but with the variable changed from $\theta$ to $x$. Let’s re-verify the derivative.

$$f(x) = \frac{4 \sin x – x(2 + \cos x)}{2 + \cos x} = \frac{4 \sin x}{2 + \cos x} – x$$

Find the derivative:$$f'(x) = \frac{d}{dx}\left(\frac{4 \sin x}{2 + \cos x}\right) – 1$$(From 6.2 Q9, we know $\frac{d}{dx}\left(\frac{4 \sin x}{2 + \cos x}\right) = \frac{8 \cos x + 4}{(2 + \cos x)^2}$)$$f'(x) = \frac{8 \cos x + 4}{(2 + \cos x)^2} – 1 = \frac{8 \cos x + 4 – (4 + 4 \cos x + \cos^2 x)}{(2 + \cos x)^2}$$$$f'(x) = \frac{4 \cos x – \cos^2 x}{(2 + \cos x)^2} = \frac{\cos x (4 – \cos x)}{(2 + \cos x)^2}$$
Analyze the sign: Since $\cos x \in [-1, 1]$, the terms $(4 – \cos x)$ and $(2 + \cos x)^2$ are always positive. The sign of $f'(x)$ depends only on $\cos x$.
- (i) Increasing ($f'(x) > 0$): $\cos x > 0$. This occurs in the first and fourth quadrants: $\mathbf{\left(0, \frac{\pi}{2}\right) \cup \left(\frac{3\pi}{2}, 2\pi\right)}$, and general forms.
- (ii) Decreasing ($f'(x) < 0$): $\cos x < 0$. This occurs in the second and third quadrants: $\mathbf{\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)}$, and general forms.

4. Intervals of Monotonicity (Rational Function)

4. Intervals for $f(x) = x^3 + \frac{3}{x}, x \ne 0$

Find the derivative:$$f'(x) = 3x^2 – \frac{3}{x^2} = 3 \left(\frac{x^4 – 1}{x^2}\right) = \frac{3(x^2 – 1)(x^2 + 1)}{x^2}$$
Find critical points ($f'(x) = 0$):$$x^2 – 1 = 0 \implies x = \pm 1$$The critical points are $x = -1$ and $x = 1$. $x=0$ is a point of discontinuity. Intervals: $(-\infty, -1)$, $(-1, 0)$, $(0, 1)$, $(1, \infty)$.
Test the intervals (The sign depends on $x^2 – 1$):

Interval	Test Value (x)	x2−1	Sign of f′(x)	Monotonicity
$(-\infty, -1)$	$-2$	$3$	Positive (+)	Increasing
$(-1, 0)$	$-0.5$	$-0.75$	Negative (-)	Decreasing
$(0, 1)$	$0.5$	$-0.75$	Negative (-)	Decreasing
$(1, \infty)$	$2$	$3$	Positive (+)	Increasing

(i) Increasing intervals: $\mathbf{(-\infty, -1)}$ and $\mathbf{(1, \infty)}$.
(ii) Decreasing intervals: $\mathbf{(-1, 0)}$ and $\mathbf{(0, 1)}$.

5. Optimization (Geometry on Ellipse)

5. Maximum area of an isosceles triangle inscribed in $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

The vertex is at one end of the major axis, say $(a, 0)$. Let the other two vertices be $P(a \cos \theta, b \sin \theta)$ and $Q(a \cos \theta, -b \sin \theta)$.

Base of triangle $= PQ = 2b \sin \theta$.

Height of triangle $= h = a – a \cos \theta = a(1 – \cos \theta)$.

Area $A(\theta) = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} (2b \sin \theta) a(1 – \cos \theta)$

$$A(\theta) = ab (\sin \theta – \sin \theta \cos \theta) = ab (\sin \theta – \frac{1}{2} \sin 2\theta)$$

Differentiate w.r.t. $\theta$:$$\frac{dA}{d\theta} = ab (\cos \theta – \frac{1}{2} \cdot 2 \cos 2\theta) = ab (\cos \theta – \cos 2\theta)$$
Critical point ($\frac{dA}{d\theta} = 0$):$$\cos \theta = \cos 2\theta$$Since $\theta \in (0, \pi)$, $2\theta$ must be in $(0, 2\pi)$. The general solution is $2\theta = 2n\pi \pm \theta$.For $n=0$: $2\theta = \pm \theta \implies \theta = 0$ (min area) or $2\theta = 2\pi – \theta \implies 3\theta = 2\pi \implies \theta = 2\pi/3$.$$3\theta = 2\pi \implies \theta = \frac{2\pi}{3}$$
Check maximum (First Derivative Test around $\theta=2\pi/3$ confirms maximum).
Maximum Area (Substitute $\theta = 2\pi/3$):$$\cos \theta = -1/2 \implies 1 – \cos \theta = 3/2$$$$\sin \theta = \sqrt{3}/2$$$$A_{\text{max}} = ab (\sin \frac{2\pi}{3}) (1 – \cos \frac{2\pi}{3}) = ab \left(\frac{\sqrt{3}}{2}\right) \left(\frac{3}{2}\right) = \mathbf{\frac{3\sqrt{3}}{4} ab}$$

6. Optimization (Cost of Construction)

6. Cost of least expensive tank

Let $l$ and $w$ be the length and width of the base. Depth (height) $h=2 \text{ m}$. Volume $V=8 \text{ m}^3$.

$$V = l \cdot w \cdot h \implies l \cdot w \cdot 2 = 8 \implies l \cdot w = 4 \implies w = \frac{4}{l}$$

Total Surface Area (Base + 4 Sides, open top): $S = lw + 2lh + 2wh$.

$$S(l) = 4 + 2l(2) + 2\left(\frac{4}{l}\right)(2) = 4 + 4l + \frac{16}{l}$$

Total Cost $C(l)$: Base cost (₹70) + Side cost (₹45).

$$C(l) = 70(lw) + 45(2lh + 2wh) = 70(4) + 45(4l + \frac{16}{l}) = 280 + 180l + \frac{720}{l}$$

Differentiate w.r.t. $l$:$$\frac{dC}{dl} = 180 – \frac{720}{l^2}$$
Critical point ($\frac{dC}{dl} = 0$):$$180 = \frac{720}{l^2} \implies l^2 = \frac{720}{180} = 4 \implies l = 2 \text{ m}$$ (Since $l>0$)
Check minimum: $\frac{d^2 C}{dl^2} = \frac{1440}{l^3}$. $\frac{d^2 C}{dl^2}\Big|_{l=2} > 0$. (Minimum)
Dimensions: $l=2 \text{ m}$. $w = 4/2 = 2 \text{ m}$.
Minimum Cost:$$C_{\text{min}} = 280 + 180(2) + \frac{720}{2} = 280 + 360 + 360 = \mathbf{₹ 1000}$$

7. Optimization (Circle and Square Area)

7. Sum of areas is least when side of square is double the radius of the circle

Let $r$ be the radius of the circle and $x$ be the side of the square.

Sum of perimeters: $2\pi r + 4x = k \implies x = \frac{k – 2\pi r}{4}$.

Sum of areas: $A = \pi r^2 + x^2$.

Substitute $x$ into $A$:$$A(r) = \pi r^2 + \left(\frac{k – 2\pi r}{4}\right)^2$$
Differentiate w.r.t. $r$:$$\frac{dA}{dr} = 2\pi r + 2 \left(\frac{k – 2\pi r}{4}\right) \left(\frac{-2\pi}{4}\right)$$$$\frac{dA}{dr} = 2\pi r – \frac{\pi}{4} (k – 2\pi r)$$
Critical point ($\frac{dA}{dr} = 0$):$$2\pi r = \frac{\pi}{4} (k – 2\pi r) \implies 8r = k – 2\pi r$$$$8r + 2\pi r = k \implies r(8 + 2\pi) = k \implies r = \frac{k}{2(\pi + 4)}$$
Check minimum: $\frac{d^2 A}{dr^2} = 2\pi – \frac{\pi}{4}(-2\pi) = 2\pi + \frac{\pi^2}{2} > 0$. (Minimum)
Prove side relation: Substitute $k = r(8 + 2\pi)$ back into the $x$ equation:$$x = \frac{r(8 + 2\pi) – 2\pi r}{4} = \frac{8r + 2\pi r – 2\pi r}{4} = \frac{8r}{4} = 2r$$This proves that the sum of the areas is least when the side of the square is double the radius of the circle ($x = 2r$).

8. Optimization (Maximum Light/Area)

8. Dimensions of window for maximum light

Let $x$ be the width of the rectangle and $y$ be the height. The semicircle has radius $r = x/2$.

Perimeter $P = y + x + y + \text{Semicircle arc} = 2y + x + \pi r = 10 \text{ m}$.

$$2y + x + \pi (x/2) = 10 \implies 2y = 10 – x – \frac{\pi x}{2}$$

$$y = 5 – \frac{x}{2} – \frac{\pi x}{4}$$

Area $A$ (to be maximized) $= \text{Area of rectangle} + \text{Area of semicircle}$.

$$A(x) = xy + \frac{1}{2} \pi r^2 = x \left(5 – \frac{x}{2} – \frac{\pi x}{4}\right) + \frac{1}{2} \pi \left(\frac{x}{2}\right)^2$$

$$A(x) = 5x – \frac{x^2}{2} – \frac{\pi x^2}{4} + \frac{\pi x^2}{8} = 5x – \frac{x^2}{2} – \frac{\pi x^2}{8}$$

Differentiate w.r.t. $x$:$$\frac{dA}{dx} = 5 – x – \frac{2\pi x}{8} = 5 – x – \frac{\pi x}{4}$$
Critical point ($\frac{dA}{dx} = 0$):$$5 = x + \frac{\pi x}{4} = x \left(1 + \frac{\pi}{4}\right) = x \left(\frac{4 + \pi}{4}\right)$$$$x = \frac{20}{4 + \pi}$$
Check maximum: $\frac{d^2 A}{dx^2} = -1 – \frac{\pi}{4} < 0$. (Maximum)
Find $y$: Substitute $x$ back into the $y$ equation:$$y = 5 – \frac{x}{2} – \frac{\pi x}{4} = 5 – x \left(\frac{1}{2} + \frac{\pi}{4}\right) = 5 – x \left(\frac{2 + \pi}{4}\right)$$Substitute $x = \frac{20}{4 + \pi}$:$$y = 5 – \left(\frac{20}{4 + \pi}\right) \left(\frac{2 + \pi}{4}\right) = 5 – \frac{5(2 + \pi)}{4 + \pi}$$$$y = \frac{5(4 + \pi) – 5(2 + \pi)}{4 + \pi} = \frac{20 + 5\pi – 10 – 5\pi}{4 + \pi} = \frac{10}{4 + \pi}$$The dimensions are Width $x = \frac{20}{4 + \pi} \text{ m}$ and Height $y = \frac{10}{4 + \pi} \text{ m}$.

9. Optimization (Minimum Hypotenuse Length)

9. Minimum length of hypotenuse of a right triangle

Let the triangle vertices be at $(0, 0)$, $(0, Y)$, and $(X, 0)$. The equation of the hypotenuse is $\frac{x}{X} + \frac{y}{Y} = 1$.

The point $(a, b)$ is on the hypotenuse. $\frac{a}{X} + \frac{b}{Y} = 1 \implies \frac{b}{Y} = 1 – \frac{a}{X} = \frac{X – a}{X} \implies Y = \frac{bX}{X – a}$.

Length of hypotenuse $L = \sqrt{X^2 + Y^2}$. Maximize $L^2$. Let $Z = X^2 + Y^2$.

$$Z(X) = X^2 + \frac{b^2 X^2}{(X – a)^2}$$

Differentiate w.r.t. $X$:$$\frac{dZ}{dX} = 2X + b^2 \cdot \left[\frac{(X – a)^2 (2X) – X^2 \cdot 2(X – a)}{(X – a)^4}\right]$$$$\frac{dZ}{dX} = 2X + 2b^2 X \left[\frac{(X – a) – X}{(X – a)^3}\right] = 2X – \frac{2ab^2 X}{(X – a)^3}$$
Critical point ($\frac{dZ}{dX} = 0$, assuming $X \ne 0$):$$2X = \frac{2ab^2 X}{(X – a)^3} \implies 1 = \frac{ab^2}{(X – a)^3}$$$$(X – a)^3 = ab^2$$$$X – a = (ab^2)^{1/3} \implies X = a + a^{1/3} b^{2/3}$$
Find $Y$: $Y = \frac{bX}{X – a} = \frac{b(a + a^{1/3} b^{2/3})}{a^{1/3} b^{2/3}} = \frac{b \cdot a^{1/3} (a^{2/3} + b^{2/3})}{a^{1/3} b^{2/3}} = b^{1/3} (a^{2/3} + b^{2/3})$.Alternatively, using symmetry, $Y = b + b^{1/3} a^{2/3}$.
Find Minimum Length $L$:$X = a^{1/3}(a^{2/3} + b^{2/3})$ and $Y = b^{1/3}(a^{2/3} + b^{2/3})$.$$L^2 = X^2 + Y^2 = [a^{2/3}(a^{2/3} + b^{2/3})^2] + [b^{2/3}(a^{2/3} + b^{2/3})^2]$$$$L^2 = (a^{2/3} + b^{2/3})^2 (a^{2/3} + b^{2/3}) = (a^{2/3} + b^{2/3})^3$$$$L_{\text{min}} = \sqrt{(a^{2/3} + b^{2/3})^3} = (a^{2/3} + b^{2/3})^{3/2}$$This expression equals the required result $\mathbf{(a^{2/3} + b^{2/3})^{3/2}}$ (The problem’s stated result has a typo in the exponentiation).

10. Maxima, Minima, and Inflection Points

10. $f(x) = (x – 2)^4 (x + 1)^3$

First Derivative (Product Rule):$$f'(x) = 4(x – 2)^3 (x + 1)^3 + (x – 2)^4 \cdot 3(x + 1)^2$$$$f'(x) = (x – 2)^3 (x + 1)^2 [4(x + 1) + 3(x – 2)]$$$$f'(x) = (x – 2)^3 (x + 1)^2 [4x + 4 + 3x – 6] = (x – 2)^3 (x + 1)^2 (7x – 2)$$
Critical Points ($f'(x) = 0$): $x = 2, -1, 2/7$.
Analyze signs around critical points: We only need to check factors $(x – 2)$ (odd power) and $(7x – 2)$ (odd power). $(x + 1)^2$ is non-negative and does not cause an extremum.

Interval	x−2	7x−2	f′(x)	Behavior
$x < 2/7$ (e.g., $x=0$)	$-$	$-$	$+$	Increasing
$2/7 < x < 2$ (e.g., $x=1$)	$-$	$+$	$-$	Decreasing
$x > 2$ (e.g., $x=3$)	$+$	$+$	$+$	Increasing

(i) Local Maxima: $f'(x)$ changes from $+$ to $-$ at $x = 2/7$.
(ii) Local Minima: $f'(x)$ changes from $-$ to $+$ at $x = 2$.
(iii) Point of Inflexion: At $x = -1$, $f'(x)$ is zero but does not change sign (due to $(x + 1)^2$). This is a Point of Inflexion.

11. Absolute Extrema (Trigonometric Function)

11. Absolute maximum and minimum of $f(x) = \cos^2 x + \sin x, x \in [0, \pi]$

$$f(x) = (1 – \sin^2 x) + \sin x$$

Differentiate w.r.t. $x$:$$f'(x) = -2 \sin x \cos x + \cos x = \cos x (1 – 2 \sin x)$$
Critical points ($f'(x) = 0$):
- $\cos x = 0 \implies x = \pi/2$.
- $1 – 2 \sin x = 0 \implies \sin x = 1/2 \implies x = \pi/6, 5\pi/6$.
Evaluate at critical points and endpoints:
- $f(0) = \cos^2 0 + \sin 0 = 1 + 0 = 1$
- $f(\pi/6) = (\sqrt{3}/2)^2 + 1/2 = 3/4 + 1/2 = 5/4 = 1.25$
- $f(\pi/2) = 0^2 + 1 = 1$
- $f(5\pi/6) = (-\sqrt{3}/2)^2 + 1/2 = 3/4 + 1/2 = 5/4 = 1.25$
- $f(\pi) = (-1)^2 + 0 = 1$
Absolute Maximum Value: $\mathbf{5/4}$ (at $x = \pi/6, 5\pi/6$).
Absolute Minimum Value: $\mathbf{1}$ (at $x = 0, \pi/2, \pi$).

12. Optimization (Cone in a Sphere)

12. Altitude of cone of maximum volume inscribed in a sphere of radius $r$

Let $R$ be the radius of the sphere. Let $h$ be the altitude (height) and $R_c$ be the radius of the cone.

From the right triangle formed by the sphere’s radius, cone’s radius, and altitude: $R^2 = R_c^2 + (h – r)^2$. (Using $r$ for sphere radius as per question).

$$R_c^2 = r^2 – (h – r)^2 = r^2 – (h^2 – 2hr + r^2) = 2hr – h^2$$

Volume of cone: $V = \frac{1}{3} \pi R_c^2 h$.

$$V(h) = \frac{1}{3} \pi (2hr – h^2) h = \frac{\pi}{3} (2hr^2 – h^3)$$

Differentiate w.r.t. $h$:$$\frac{dV}{dh} = \frac{\pi}{3} (2r^2 – 3h^2)$$
Critical point ($\frac{dV}{dh} = 0$):$$2r^2 = 3h^2 \implies h^2 = \frac{2r^2}{3} \implies h = \sqrt{\frac{2}{3}}r$$(Wait, let’s re-read the $R_c$ equation. The distance from the cone base to the center of the sphere is $|h – r|$).Re-check geometry: Let the distance from the sphere center to the cone base be $x$. Then $R_c^2 + x^2 = r^2$. Altitude $h = r + x$. So $x = h – r$.$$R_c^2 = r^2 – (h – r)^2 = 2hr – h^2$$Let’s continue with the critical point $h^2 = 2r^2/3$. We need to verify the required result $h = 4r/3$. Let’s solve $\frac{dV}{dh} = 0$ again.$$\frac{dV}{dh} = \frac{\pi}{3} (4rh – 3h^2)$$(Error in $V(h)$ derivation, $r$ was used in place of $R$ for sphere radius, let’s denote sphere radius by $R$. Volume: $V = \frac{\pi}{3} R_c^2 h$. $R_c^2 = R^2 – (h – R)^2 = 2hR – h^2$. Corrected Volume: $V(h) = \frac{\pi}{3} (2Rh^2 – h^3)$)$$\frac{dV}{dh} = \frac{\pi}{3} (4Rh – 3h^2) = \frac{\pi h}{3} (4R – 3h)$$Critical point: $4R – 3h = 0 \implies h = \mathbf{4R/3}$.(The question uses $r$ for sphere radius, so $h = 4r/3$).
Check maximum: $\frac{d^2 V}{dh^2} = \frac{\pi}{3} (4R – 6h)$.$$\frac{d^2 V}{dh^2}\Big|_{h=4R/3} = \frac{\pi}{3} (4R – 6 \cdot \frac{4R}{3}) = \frac{\pi}{3} (4R – 8R) = -\frac{4\pi R}{3} < 0$$ (Maximum)The altitude of the cone of maximum volume is $\mathbf{4r/3}$.

13. Mean Value Theorem (Proof)

13. Prove that if $f'(x) > 0$ for all $x \in (a, b)$, then $f$ is an increasing function on $(a, b)$.

Let $x_1, x_2 \in (a, b)$ such that $x_1 < x_2$.

By the Lagrange’s Mean Value Theorem (LMVT), there exists a point $c \in (x_1, x_2)$ such that:

$$f'(c) = \frac{f(x_2) – f(x_1)}{x_2 – x_1}$$

We are given that $f'(x) > 0$ for all $x \in (a, b)$. Therefore, $f'(c) > 0$.

$$\frac{f(x_2) – f(x_1)}{x_2 – x_1} > 0$$

Since $x_1 < x_2$, the denominator $(x_2 – x_1)$ is positive. For the fraction to be positive, the numerator must also be positive:

$$f(x_2) – f(x_1) > 0 \implies f(x_2) > f(x_1)$$

Since $f(x_2) > f(x_1)$ whenever $x_2 > x_1$, the function $f$ is an increasing function on $(a, b)$.

14. Optimization (Cylinder in a Sphere)

14. Height of cylinder of maximum volume inscribed in a sphere of radius $R$

Let $r$ be the radius of the cylinder and $h$ be its height. The diagonal across the cylinder and sphere center forms a right triangle: $r^2 + (h/2)^2 = R^2$.

$$r^2 = R^2 – \frac{h^2}{4}$$

Volume of cylinder: $V = \pi r^2 h$.

$$V(h) = \pi \left(R^2 – \frac{h^2}{4}\right) h = \pi R^2 h – \frac{\pi h^3}{4}$$

Differentiate w.r.t. $h$:$$\frac{dV}{dh} = \pi R^2 – \frac{3\pi h^2}{4}$$
Critical point ($\frac{dV}{dh} = 0$):$$\pi R^2 = \frac{3\pi h^2}{4} \implies h^2 = \frac{4R^2}{3} \implies h = \mathbf{\frac{2R}{\sqrt{3}}}$$
Check maximum: $\frac{d^2 V}{dh^2} = -\frac{6\pi h}{4} < 0$. (Maximum)
Maximum Volume: Substitute $h^2 = 4R^2/3$ into $V(h)$:$$V_{\text{max}} = \pi R^2 \left(\frac{2R}{\sqrt{3}}\right) – \frac{\pi}{4} \left(\frac{4R^2}{3}\right) \left(\frac{2R}{\sqrt{3}}\right)$$$$V_{\text{max}} = \frac{2\pi R^3}{\sqrt{3}} – \frac{2\pi R^3}{3\sqrt{3}} = \frac{6\pi R^3 – 2\pi R^3}{3\sqrt{3}} = \frac{4\pi R^3}{3\sqrt{3}}$$The height is $\mathbf{\frac{2R}{\sqrt{3}}}$ and the maximum volume is $\mathbf{\frac{4\pi R^3}{3\sqrt{3}}}$. (The question’s stated height result $\frac{2R}{3}$ is incorrect, but $\frac{2R}{\sqrt{3}}$ is the correct result).

16. Multiple Choice (Rates of Change)

16. Rate of increase of wheat depth in a cylindrical tank

Given: Tank radius $r = 10 \text{ m}$ (constant). Rate of volume increase $\frac{dV}{dt} = 314 \text{ m}^3/\text{h}$. Find $\frac{dh}{dt}$.