Exercise-6.3, Class 12th, Maths, Chapter 6, NCERT

Question 11.
It is given that at $x=1$, the function

$$f(x)=x^4-62x^2+ax+9$$

attains its maximum value on the interval $[0,2]$. Find the value of $a$

.Solution

Since the function attains a maximum at $x=1$

 (an interior point of the interval $[0,2]$),
the necessary condition for maxima (from derivative test) is:

$$f^{\mathrm{\prime }}(1)=0$$

First, differentiate the function:

$$f(x)=x^4-62x^2+ax+9$$

$$f^{\mathrm{\prime }}(x)=4x^3-124x+a$$

Put $x=\mathrm{1 }$and set $f^{\mathrm{\prime }}(1)=0$:

$$f^{\mathrm{\prime }}(1)=4(1{)}^3-124(1)+a=0$$

$$4-124+a=0$$

$$a-120=0$$

$$\boxed{{\displaystyle a=120}}$$

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Question 12.
Find the maximum and minimum values of the function

$$f(x)=x+\sin 2x\text{on the interval }[0,2\pi ]\mathrm{.}$$

Solution

Let$$f(x)=x+\sin 2x$$

Step 1: Find the derivative

$$f^{\mathrm{\prime }}(x)=\frac{d}{dx}(x)+\frac{d}{dx}(\sin 2x)$$

$$f^{\mathrm{\prime }}(x)=1+2\cos 2x$$

Step 2: Put $f^{\mathrm{\prime }}(x)=\mathrm{0 }$to find critical points

$$1+2\cos 2x=0$$

$$2\cos 2x=-1$$

$$\cos 2x=-\frac{1}{2}$$

Step 3: Solve for $x$

$$2x=\frac{2\pi }{3},\frac{4\pi }{3},\frac{8\pi }{3},\frac{10\pi }{3}$$

Dividing by 2:$$x=\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3}$$

These points lie inside $[0,2\pi ]$

.Step 4: Evaluate $f(x)$at critical points and endpoints

Compute $f(x)=x+\sin 2x$:

Step 5: Identify maximum and minimum

Comparing the values:

• Largest value occurs at
  $x=\frac{4\pi }{3}$:

$$f_{\max }=\frac{4\pi }{3}+\frac{\sqrt{3}}{2}$$

• Smallest value occurs at
  $x=\mathrm{0<span\; style="font-family:\; var(--wp--preset--font-family--manrope);\; font-size:\; var(--wp--preset--font-size--large);\; letter-spacing:\; -0.1px;">:</span>}$

$$f_{\min }=0$$Final Answer

$$\boxed{{\displaystyle \text{Maximum value }=\frac{4\pi }{3}+\frac{\sqrt{3}}{2}}}$$

$$\boxed{{\displaystyle \text{Minimum value }=0}}$$

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Question 13.
Find two numbers whose sum is 24 and whose product is as large as possible.

Solution

Let the two numbers be $x$ and $y$.
Given:

$$x+y=24\Rightarrow y=24-x$$

We want to maximize the product:

$$P=x\cdot y=x(24-x)=24x-x^2$$So,

$$P(x)=24x-x^2$$

Step 1: Differentiate

$$P^{\mathrm{\prime }}(x)=24-2x$$

Step 2: Set $P^{\mathrm{\prime }}(x)=0$

$$24-2x=0$$

$$2x=24$$

$$x=12$$

Step 3: Find the corresponding second number

$$y=24-x=24-12=12$$

Step 4: Second derivative test

$$P^{\mathrm{\prime }\mathrm{\prime }}(x)=-2<0$$

Since $P^{\mathrm{\prime }\mathrm{\prime }}(x)<0$, $\mathrm{P }$has a maximum at $x=12$.

​​Their product is maximum when both are equal.​

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Question 14.
Find two positive numbers $x$ and $\mathrm{y }$such that $x+y=\mathrm{60 }$and $x{y}^{\mathrm{3 }}$is maximum.

Solution

Let the required expression be:

$$P=x{y}^3$$Given:

$$x+y=60\Rightarrow x=60-y$$

Substitute into $P$:

$$P(y)=(60-y)y^3=60y^3-y^4$$

Step 1: Differentiate

$$P^{\mathrm{\prime }}(y)=180y^2-4y^3=4y^2(45-y)$$

Step 2: Find critical points

Set $P^{\mathrm{\prime }}(y)=0$

$$4y^2(45-y)=0$$So,

$$y=0,y=45$$

Since numbers are positive, we consider only:

$$y=45$$

Then,

$$x=60-y=60-45=15$$

Step 3: Second derivative test

$$P^{\mathrm{\prime }\mathrm{\prime }}(y)=360y-12y^2$$

$$P^{\mathrm{\prime }\mathrm{\prime }}(45)=360(45)-12(45{)}^2=16200-24300=-8100<0$$

Since $P^{\mathrm{\prime }\mathrm{\prime }}(45)<0$, the function has a maximum at $y=45$

.Final Answer

$$\boxed{{\displaystyle x=15,y=45}}$$

$$\boxed{{\displaystyle Th\mathrm{e }product}}x{y}^3\text{ is maximum when }x=15\text{ and }y=45.$$

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Question 15.
Find two positive numbers $\mathrm{x }$and $\mathrm{y }$such that their sum is 35 and the product $x^2{y}^{\mathrm{5 }}$is maximum.

Solution

Let:

$$P=x^2{y}^5$$Given:

$$x+y=35\Rightarrow x=35-y$$

Substitute in product:

$$P(y)=(35-y{)}^2{y}^5$$

Step 1: Take logarithm for easier differentiation

$$\ln P=\ln ((35-y{)}^2{y}^5)$$

$$\ln P=2\ln (35-y)+5\ln y$$

Differentiate both sides w.r.t.$y$:

$$\frac{1}{P}\cdot P^{\mathrm{\prime }}=2\cdot \frac{-1}{35-y}+5\cdot \frac{1}{y}$$So,

$$P^{\mathrm{\prime }}=P(\frac{-2}{35-y}+\frac{5}{y})$$Set

$P^{\mathrm{\prime }}=0$:

$$\frac{-2}{35-y}+\frac{5}{y}=0$$

Step 2: Solve the equation

$$\frac{5}{y}=\frac{2}{35-y}$$Cross multiply:

$$5(35-y)=2y$$

$$175-5y=2y$$

$$175=7y$$

$$y=25$$Now,$$x=35-25=10$$

Step 3: Second derivative test

(Since $P^{\mathrm{\prime }}=0$ yields a maximum in similar problems with positive product forms, we conclude maximum)

Final Answer

$$\boxed{{\displaystyle x=10,y=25}}$$

$$\boxed{{\displaystyle \text{The product }{x}^2{y}^5\text{ is maximum when }x=10\text{ and }y=25.}}$$

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Question 16.
Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Solution

Let the two positive numbers be $x$ and $y$.
Given:$$x+y=16\Rightarrow y=16-x$$

We want to minimise:

$$S=x^3+y^3=x^3+(16-x{)}^3$$

Step 1: Write $\mathrm{S }$in terms of $x$

$$S(x)=x^3+(16-x{)}^3$$

Expand:

$$S(x)=x^3+(4096-768x+48x^2-x^3)$$

$$S(x)=4096-768x+48x^2$$So,$$S^{\mathrm{\prime }}(x)=96x-768$$Step 2: Set $S^{\mathrm{\prime }}(x)=0$

$$96x-768=0$$

$$96x=768$$

$$x=8$$Then,$$y=16-8=8$$

Step 3: Second derivative test

$$S^{\mathrm{\prime }\mathrm{\prime }}(x)=96>0$$

Since $S^{\mathrm{\prime }\mathrm{\prime }}(x)>0$, the value at $x=8$ is a minimum.

Final Answer

$$\boxed{{\displaystyle x=8\text{ and }y=8}}$$

$$\boxed{{\displaystyle \text{The sum of the cubes is minimum when the two numbers are equal.}}}$$

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Question 17

A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps. What should be the side length of the square cut off so that the volume of the box is maximum? Also give the image.

Solution

Let the side of the square cut from each corner be $x$ cm.

When folded, the resulting box has:

• Height = $x$

• Length = $18-2x$

• Width = $18-2x$

So the volume $V$ of the open box:

$$V=x(18-2x{)}^2$$Step 1: Expand

$$V=x(324-72x+4x^2)=324x-72x^2+4x^3$$

Step 2: Differentiate

$$V^{\mathrm{\prime }}=324-144x+12x^2$$

Set $V^{\mathrm{\prime }}=0$:

$$12x^2-144x+324=0$$

Divide by 12:$$x^2-12x+27=0$$

Step 3: Solve quadratic

$$x=\frac{12\pm \sqrt{144-108}}{2}=\frac{12\pm \sqrt{36}}{2}=\frac{12\pm 6}{2}$$So,

$$x=9\text{or}x=3$$

Since the box must have positive dimensions and

$18-2x>0$, $x=\mathrm{9 }$gives zero base, so we reject x = 9.

$$x=3\text{ cm}$$

Step 4: Second derivative test

$$V^{\mathrm{\prime }\mathrm{\prime }}=24x-144$$

$$V^{\mathrm{\prime }\mathrm{\prime }}(3)=72-144=-72<0$$

So $x=3$ cm gives a maximum volume.

Final Answer

$$\boxed{{\displaystyle x=3\text{ cm}}}$$

The side of the square to be cut off must be 3 cm.

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Question 18

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off a square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?

Solution

• Let the side of the square cut from each corner be $x$ cm.

  After cutting and folding:

  • Height of the box = $x$

  • Length of base = $45-2x$

  • Width of base = $24-2x$

  
  $$$$

So the volume of the open box:

$$V=x(45-2x)(24-2x)$$

Step 1: Expand

$$V=x(45-2x)(24-2x)$$

$$=x(1080-90x-48x+4x^2)$$

$$=x(1080-138x+4x^2)$$

$$=1080x-138x^2+4x^3$$

Step 2: Differentiate w.r.t. $x$

$$V^{\mathrm{\prime }}=1080-276x+12x^2$$

Set $V^{\mathrm{\prime }}=0$:

$$12x^2-276x+1080=0$$

Divide by 12:

$$x^2-23x+90=0$$

Step 3: Solve using quadratic formula

$$x=\frac{23\pm \sqrt{{23}^2-4\cdot 90}}{2}$$

$$=\frac{23\pm \sqrt{529-360}}{2}$$

$$=\frac{23\pm \sqrt{169}}{2}$$

$$=\frac{23\pm 13}{2}$$So,

$$x=\frac{23+13}{2}=18\text{(not possible because width becomes negative)}$$

$$x=\frac{23-13}{2}=5$$

Step 4: Second derivative test

$$V^{\mathrm{\prime }\mathrm{\prime }}=24x-276$$

$$V^{\mathrm{\prime }\mathrm{\prime }}(5)=120-276=-156<0$$

So $x=5$ cm gives maximum volume.

Final Answer

$$\boxed{{\displaystyle x=5\text{ cm}}}$$

The side of the square to be cut off must be 5 cm.

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Question 19

Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Solution

Let a rectangle $ABCD$ be inscribed in a circle of radius $r$.

Let:

• Half the length of the rectangle = $x$

• Half the breadth of the rectangle = $y$

Then full dimensions = $2x\times 2y$

The diagonal of the rectangle equals the diameter of the circle:

$$(2x{)}^2+(2y{)}^2=(2r{)}^2$$

$$4x^2+4y^2=4r^2$$

$$x^2+y^2=r^2\text{(1)}$$

Area of the rectangle

$$A=\text{length}\times \text{breadth}=(2x)(2y)=4xy$$

To maximize area, we maximize the product $xy$.

From (1):

$$y=\sqrt{r^2-x^2}$$

So,$$A(x)=4x\sqrt{r^2-x^2}$$

Differentiate

Let

$$A=4x(r^2-x^2{)}^{1\mathrm{/}2}$$
$$A^{\mathrm{\prime }}=4[\sqrt{r^2-x^2}-\frac{x^2}{\sqrt{r^2-x^2}}]$$
$$A^{\mathrm{\prime }}=\frac{4(r^2-2x^2)}{\sqrt{r^2-x^2}}$$

Set $A^{\mathrm{\prime }}=0$:

$$r^2-2x^2=0\Rightarrow x^2=\frac{r^2}{2}$$

Thus,$$x=y=\frac{r}{\sqrt{2}}$$

So, the rectangle becomes a square.

Final Conclusion

$$\boxed{{\displaystyle \text{The area is maximum when the rectangle is a square.}}}$$

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Question 20

Show that the right circular cylinder of given surface area and maximum volume is such that its height is equal to the diameter of the base.

Solution

Let the radius of the cylindrical base be $r$ and height be $h$.

Surface area constraint

The total surface area $S$ of a closed cylinder is:

$$S=2\pi r^2+2\pi rh$$

(Since surface is given and fixed, it is a constant.)

Volume of cylinder

$$V=\pi r^{2}h$$

Using the surface constraint, solve for $h$:

$$2\pi r^2+2\pi rh=S$$
$$2\pi rh=S-2\pi r^2$$
$$h=\frac{S-2\pi r^2}{2\pi r}$$

$$h=\frac{S}{2\pi r}-r$$

Substitute into volume:

$$V(r)=\pi r^2(\frac{S}{2\pi r}-r)$$

$$V(r)=\frac{Sr}{2}-\pi r^3$$

Differentiate for maximum

$$V^{\mathrm{\prime }}(r)=\frac{S}{2}-3\pi r^2$$

Set $V^{\mathrm{\prime }}(r)=0$:

$$\frac{S}{2}=3\pi r^2$$

$$r^2=\frac{S}{6\pi }$$

Find $h$

$$h=\frac{S}{2\pi r}-r=\frac{S}{2\pi \sqrt{S\mathrm{/}(6\pi )}}-\sqrt{S\mathrm{/}(6\pi )}$$

Simplifying, we get:

$$h=2r$$

Final Result

$$\boxed{{\displaystyle \text{For maximum volume, the height }h\text{ of the cylinder must be equal to the diameter }2r\mathrm{.}}}$$
$$\boxed{{\displaystyle h=2r}}$$