NCERT Exemplar Solutions Class 12 Maths Application of Derivatives — Long Answer Q25–34

NCERT Exemplar Problems, Class 12 Mathematics — Application of Derivatives, Long Answer questions 25 to 34. All 10 questions solved.

These ten questions are the heart of the chapter: write the quantity to be maximised or minimised, reduce it to one variable using the given constraint, set the derivative to zero, and then use the second derivative (or a sign change of the first) to confirm which kind of turning point you have. Question 26 asks for local extrema and points of inflection of a quintic, and Question 28 is a tangency proof.

A note on the syllabus. The rationalised NCERT Chapter 6 keeps only rate of change, increasing and decreasing functions, and maxima and minima (Exercises 6.1 to 6.3 and the Miscellaneous Exercise). It dropped tangents and normals, and approximations. JEE Main 2026 lists only rate of change, monotonicity and maxima and minima. JEE Advanced 2026 keeps “tangents and normals” but does not list approximations. On this page only Question 28 (a tangency condition) falls under tangents and normals, so it is outside the current textbook and Main but relevant to Advanced. Questions 25 to 27 and 29 to 34 are all maxima and minima and are fully in syllabus. Points of inflection in Question 26 are a standard part of curve study, though the textbook now stresses only local extrema.

Key insight. Eliminate until one variable remains, solve the first derivative equal to zero, and confirm the nature of the point with the second derivative. For constrained problems, differentiating the constraint implicitly (Questions 31 and 34) is often cleaner than solving for one variable.

Question 25

If the sum of the lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum when the angle between them is $\dfrac{\pi}{3}$.

Solution. Let the side be $x$ and the hypotenuse $h$, with $x+h=k$ fixed. The third side is $\sqrt{h^2-x^2}=\sqrt{(k-x)^2-x^2}=\sqrt{k^2-2kx}$, so the area is $A=\tfrac12x\sqrt{k^2-2kx}$. Maximising $A$ is the same as maximising $A^2$, which avoids the root: $A^2=\dfrac{x^2}{4}(k^2-2kx)=\dfrac{k^2x^2-2kx^3}{4}$. Then $\dfrac{d(A^2)}{dx}=\dfrac{2k^2x-6kx^2}{4}=\dfrac{kx(k-3x)}{2}$, which is positive for $x<\tfrac k3$ and negative for $x>\tfrac k3$, so $A^2$ (and hence $A$) is maximum at $x=\dfrac k3$. Then $h=k-x=\dfrac{2k}{3}$, and the angle $\theta$ between the side and the hypotenuse satisfies $\cos\theta=\dfrac xh=\dfrac12$.

Question 26

Find the points of local maxima, local minima and the points of inflection of the function $f(x)=x^5-5x^4+5x^3-1$. Also find the corresponding local maximum and local minimum values.

Solution. Differentiate: $f'(x)=5x^4-20x^3+15x^2=5x^2(x^2-4x+3)=5x^2(x-1)(x-3)$. The critical points are $x=0,1,3$. Check the sign of $f’$ on each side. For $x<1$ (except $x=0$) the factor $(x-1)(x-3)$ is positive, so $f’>0$. On $(1,3)$ it is negative, and for $x>3$ it is positive. So $f’$ changes from $+$ to $-$ at $x=1$ (a local maximum) and from $-$ to $+$ at $x=3$ (a local minimum). At $x=0$ the factor $x^2$ does not change sign, so $f’$ stays positive on both sides and $x=0$ is neither a maximum nor a minimum. The values are $f(1)=1-5+5-1=0$ and $f(3)=243-405+135-1=-28$.

For inflection points, look where the concavity changes: $f”(x)=20x^3-60x^2+30x=10x(2x^2-6x+3)$. The roots are $x=0$ and $x=\dfrac{3\pm\sqrt3}{2}$. All three are simple roots of $f”$, so $f”$ changes sign at each, and each is a point of inflection. (At $x=0$ we also have $f'(0)=0$ without a sign change, so it is a stationary point of inflection.) The printed answer key gives only $x=0$ as a point of inflection; that is incomplete, because $f”$ also changes sign at $x=\dfrac{3-\sqrt3}{2}$ and $x=\dfrac{3+\sqrt3}{2}$.

Question 27

A telephone company in a town has $500$ subscribers on its list and collects fixed charges of Rs $300$/- per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of Re $1$/- one subscriber will discontinue the service. Find what increase will bring maximum profit?

Solution. Let the increase be Rs $x$. Then the charge per subscriber is $300+x$ and the number of subscribers is $500-x$, so the annual revenue is $R(x)=(500-x)(300+x)$. Differentiating, $R'(x)=-(300+x)+(500-x)=200-2x$, which is zero at $x=100$. Since $R”(x)=-2<0$, this is a maximum. The revenue is then $400\times400=\text{Rs }1{,}60{,}000$, compared with $500\times300=\text{Rs }1{,}50{,}000$ at present.

Question 28

If the straight line $x\cos\alpha+y\sin\alpha=p$ touches the curve $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, then prove that $a^2\cos^2\alpha+b^2\sin^2\alpha=p^2$.

Solution. The tangent to the ellipse at the point $(a\cos t,\,b\sin t)$ is $\dfrac{x\cos t}{a}+\dfrac{y\sin t}{b}=1$. If the given line touches the ellipse, it is this tangent for some $t$. Write the given line as $\dfrac{\cos\alpha}{p}\,x+\dfrac{\sin\alpha}{p}\,y=1$, with $p\ne0$ (a line through the centre cannot touch the ellipse). Two lines with the same right-hand side $1$ coincide when their coefficients match: $\dfrac{\cos t}{a}=\dfrac{\cos\alpha}{p}$ and $\dfrac{\sin t}{b}=\dfrac{\sin\alpha}{p}$. So $\cos t=\dfrac{a\cos\alpha}{p}$ and $\sin t=\dfrac{b\sin\alpha}{p}$. Squaring and adding, using $\cos^2t+\sin^2t=1$, gives $\dfrac{a^2\cos^2\alpha+b^2\sin^2\alpha}{p^2}=1$.

Question 29

An open box with square base is to be made of a given quantity of card board of area $c^2$. Show that the maximum volume of the box is $\dfrac{c^3}{6\sqrt3}$ cubic units.

Solution. Let the base have side $x$ and the box height $h$. The card board covers one base and four sides, so $x^2+4xh=c^2$, giving $h=\dfrac{c^2-x^2}{4x}$. The volume is $V=x^2h=\dfrac{x(c^2-x^2)}{4}=\dfrac{c^2x-x^3}{4}$. Then $V’=\dfrac{c^2-3x^2}{4}=0$ gives $x=\dfrac{c}{\sqrt3}$ (taking the positive root), and $V”=-\dfrac{6x}{4}<0$, so this is a maximum. Substituting, $V=\dfrac{1}{4}\cdot\dfrac{c}{\sqrt3}\left(c^2-\dfrac{c^2}{3}\right)=\dfrac14\cdot\dfrac{c}{\sqrt3}\cdot\dfrac{2c^2}{3}=\dfrac{c^3}{6\sqrt3}$.

Question 30

Find the dimensions of the rectangle of perimeter $36$ cm which will sweep out a volume as large as possible, when revolved about one of its sides. Also find the maximum volume.

Solution. Let the sides be $l$ and $b$, so $l+b=18$. Revolving about the side $b$ sweeps a cylinder of radius $l=18-b$ and height $b$, with volume $V=\pi(18-b)^2b$. Then $V’=\pi\left[(18-b)^2-2b(18-b)\right]=\pi(18-b)(18-3b)$, which is zero at $b=6$ (the root $b=18$ gives zero volume). Also $V”=\pi(6b-72)$, which is $-36\pi<0$ at $b=6$, so this is a maximum. The rectangle is therefore $12$ cm by $6$ cm, revolved about the $6$ cm side, so that the radius is $12$ and the height $6$: $V=\pi\cdot144\cdot6=864\pi$ cm$^3$.

Question 31

If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?

Solution. Let the cube have edge $a$ and the sphere radius $r$. The constraint is $6a^2+4\pi r^2=K$ (constant) and the quantity to minimise is $V=a^3+\tfrac43\pi r^3$. Differentiate the constraint with respect to $a$: $12a+8\pi r\dfrac{dr}{da}=0$, so $\dfrac{dr}{da}=-\dfrac{3a}{2\pi r}$. Then $\dfrac{dV}{da}=3a^2+4\pi r^2\dfrac{dr}{da}=3a^2-6ar=3a(a-2r)$. This is zero when $a=2r$, which is the diameter of the sphere. It changes from negative to positive as $a$ increases through $2r$ (and $r$ decreases with $a$), so the sum of the volumes is a minimum there. The ratio of the edge of the cube to the diameter of the sphere is $a:2r=1:1$.

Question 32

$AB$ is a diameter of a circle and $C$ is any point on the circle. Show that the area of $\Delta ABC$ is maximum, when it is isosceles.

Solution. An angle in a semicircle is a right angle, so $\angle ACB=90^\circ$. Let $AC=x$ and $BC=y$. Then $x^2+y^2=AB^2=4r^2$ is fixed, where $r$ is the radius, and the area is $\tfrac12xy$. Maximise the square, $\tfrac14x^2y^2=\tfrac14x^2(4r^2-x^2)$. Put $u=x^2$: $g(u)=u(4r^2-u)$ has $g'(u)=4r^2-2u=0$ at $u=2r^2$, and $g”=-2<0$, so it is a maximum. Then $x^2=2r^2$ and $y^2=4r^2-2r^2=2r^2$, so $AC=BC$ and the triangle is isosceles. The maximum area is $\tfrac12\cdot2r^2=r^2$.

Question 33

A metal box with a square base and vertical sides is to contain $1024$ cm$^3$. The material for the top and bottom costs Rs $5$/cm$^2$ and the material for the sides costs Rs $2.50$/cm$^2$. Find the least cost of the box.

Solution. Let the base side be $x$ cm and the height $h$ cm, so $x^2h=1024$ and $h=\dfrac{1024}{x^2}$. The top and bottom have total area $2x^2$, costing $10x^2$. The four sides have area $4xh$, costing $2.5\times4xh=10xh=\dfrac{10240}{x}$. So the cost is $C(x)=10x^2+\dfrac{10240}{x}$. Then $C'(x)=20x-\dfrac{10240}{x^2}=0$ gives $x^3=512$, so $x=8$. The second derivative $C”(x)=20+\dfrac{20480}{x^3}>0$, so this is a minimum. Then $h=\dfrac{1024}{64}=16$ and $C(8)=640+1280=1920$.

Question 34

The sum of the surface areas of a rectangular parallelopiped with sides $x$, $2x$ and $\dfrac{x}{3}$ and a sphere is given to be constant. Prove that the sum of their volumes is minimum, if $x$ is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes.

Solution. The surface area of the box is $2\left(x\cdot2x+x\cdot\dfrac x3+2x\cdot\dfrac x3\right)$. The three distinct face areas are $2x^2$, $\dfrac{x^2}{3}$ and $\dfrac{2x^2}{3}$, which add to $3x^2$, so the surface area is $6x^2$. Let the sphere have radius $r$. The constraint is $6x^2+4\pi r^2=K$ (constant), and the sum of the volumes is $V=x\cdot2x\cdot\dfrac x3+\dfrac43\pi r^3=\dfrac23x^3+\dfrac43\pi r^3$. Differentiating the constraint, $12x+8\pi r\dfrac{dr}{dx}=0$, so $\dfrac{dr}{dx}=-\dfrac{3x}{2\pi r}$. Then $\dfrac{dV}{dx}=2x^2+4\pi r^2\dfrac{dr}{dx}=2x^2-6xr=2x(x-3r)$, which is zero when $x=3r$. For the nature of the point, $\dfrac{d^2V}{dx^2}=4x-6r-6x\dfrac{dr}{dx}=4x-6r+\dfrac{9x^2}{\pi r}$, which at $x=3r$ equals $6r+\dfrac{81r}{\pi}>0$, so $V$ is a minimum. Substituting $r=\dfrac x3$: $V_{\min}=\dfrac23x^3+\dfrac{4\pi}{3}\cdot\dfrac{x^3}{27}=\dfrac23x^3\left(1+\dfrac{2\pi}{27}\right)$, which in terms of $r$ is $18r^3+\dfrac43\pi r^3$.

Common mistakes

• Treating every zero of $f’$ as an extremum. In Question 26, $f'(0)=0$ but $f’$ does not change sign at $x=0$ because of the squared factor $x^2$, so it is neither a maximum nor a minimum.
• Naming only one point of inflection. A point of inflection needs $f”$ to change sign, not just $f”=0$ at one root; all three roots of $f”=10x(2x^2-6x+3)$ qualify in Question 26.
• Setting up the wrong surface area. In Question 29 the box is open, so there is one base and four sides, giving $x^2+4xh$, not $2x^2+4xh$.
• Not confirming the nature of the turning point. Setting $V’=0$ finds a candidate only; check $V”$ or the sign change, as in Questions 30 and 34, because a maximum and a minimum look the same at that stage.
• Revolving about the wrong side in Question 30. The solid’s radius is the side perpendicular to the axis, so the axis side becomes the height; swapping them changes the cubic and the answer.

Practise next

• Exemplar Short Answer Q1–24, rate of change, tangents, monotonicity and a few maxima problems.
• Exemplar Objective and Fill in the Blanks Q35–64, the next questions.
• NCERT textbook Exercise 6.1 for rate of change.
• NCERT textbook Exercise 6.2 for increasing and decreasing functions.
• NCERT textbook Exercise 6.3 for maxima and minima, as in Questions 25 and 27 to 34.
• NCERT textbook Miscellaneous Exercise for further optimisation problems.