Mathematics Part Ii Chapter 8 Application Of Integrals

Maximize Z  =  0.1x + 0.09 y
x + y ≤ 50000
x  ≥ 20000
y  ≥ 10000
y ≤ x

Consider the vertices, A(-1, 2), B(1, 5) and C(3, 4).
Let us find the equation of the sides of the triangle 
Thus, the equation of AB is:


Now the area of  = Area of  + Area of 

The shaded area OBAO represents the area bounded by the curve x² = 4y and the line x = 4y – 2.

Let A and B be the points of intersection of the line and parabola.
Co-ordinates of point A are  Co-ordinates of point B are (2, 1).
Area OBAO = Area OBCO + Area OACO   ...(1)
Area OBCO = 

Area OACO = 
  
Therefore, required area = 

Given equations of the circles are:

Equation (1) is a circle with centre O at the origin and radius 2. Equation (2) is a circle with centre C(2, 0) and radius 2.
Solving (1) and (2), we have:

This gives 
Thus, the points of intersection of the given circles are 

Required area
 = Area of the region OACA'O
 = 2[area of the region ODCAO] 
 =2[area of the region ODAO + area of the region DCAD]

suppose a number of necklaces manufacture be x,
and the number of bracelets manufactures by y.
since the total number of items are at most 24.
x+y ≤24 .... (1)
Bracelets takes 1 hour to manufacture and necklaces take half an hour to manufacture.
x items take x hour to manufacture and y items take y/2 hour to manufacture.
and maximum time available is 16 hours.
therefore,
x/2 + y ≤ 16 ... (2)
The profit on one necklace is Rs. 100 and the profit on one bracelet is Rs. 300.
Let the profit be Z. Now we wish to maximize the profit. so,
Max z = 100 x +300y.... (3)
So, x+y ≤24
x/2 + y ≤ 16
Max Z = 100 x + 300 y is the required L.P.P

The corner points of the feasible region are A(60,0), B (120,0), C(60,30), and D (40,20).
The vaules of Z at these corner points are as follows.

The corner points of the feasible region are A(60,0), B (120,0), C(60,30), and D (40,20).
The values of Z at these corner points are as follows.

Z = 5x +10y
Z|_A(60,0) =300
Z|_B(120,0) = 600
Z|_C(60,30) =600
Z|_D(40,20) =400
Minimum value of Z = 300 at x = 60, y =0

Put y = x in  x² + y² =32

∴ x² + x² = 32

2x² = 32

x² =16

x = 4

$$\begin{gathered} \mathrm{A} = {\int }_0^4 {\mathrm{y}}_{\mathrm{line}} \mathrm{dx} + {\int }_4^{\sqrt{32}}{\mathrm{y}}_{\mathrm{circle}} \mathrm{dx} \\ \mathrm{A} = {\int }_0^4 \mathrm{x} \mathrm{dx} + {\int }_4^{\sqrt{32}} (\sqrt{32 -{\mathrm{x}}^2}\mathrm{dx} \\ = {\left(\frac{{\mathrm{x}}^2}{2}\right)}_0^4 + {\int }_4^{\sqrt{32}} (\sqrt{\sqrt{32{)}^2} - {\mathrm{x}}^2 \mathrm{dx}} \\ = \left(8\right) + {\left(\frac{\mathrm{x}}{2}\sqrt{32 - {\mathrm{x}}^2} + \frac{32}{2} {\sin }^{-1}\left(\frac{\mathrm{x}}{\sqrt{32}}\right)\right)}^{\sqrt{32}} \\ = \left(8\right) + \left(0 + 16 \mathrm{x} \frac{\mathrm{\pi }}{2} - \left(2\sqrt{16} + 16 {\sin }^{-1}\left(\frac{4}{\sqrt{32}}\right)\right)\right) \\ 8 +8\mathrm{\pi } - 8 - 16 {\sin }^{-1} \left(\frac{1}{\sqrt{2}}\right) \\ = 8\mathrm{\pi } -16 \mathrm{x} \frac{\mathrm{\pi }}{4} = 8\mathrm{\pi }- 4\mathrm{\pi } = 4\mathrm{\pi } \mathrm{sq} \mathrm{units} \end{gathered}$$

The respective equations for the parabola and the circle are:

$$\begin{gathered} {\mathrm{y}}^2 = 4\mathrm{x} ............\left(1\right) \\ 4{\mathrm{x}}^2 + 4{\mathrm{y}}^2 = 9 ............\left(2\right) \\ \mathrm{or} {\mathrm{x}}^2 + {\mathrm{y}}^2 ={\left(\frac{3}{2}\right)}^2 \end{gathered}$$

Equations (1) is a parabola with vertex ( 0, 0 ) which opens to the right and  equation (2) is a circle with centre (0, 0 ) and radius $\frac{3}{2}$.

From equations (1) and (2), we get:

$$\begin{gathered} 4{\mathrm{x}}^2 + 4 ( 4\mathrm{x} ) = 9 \\ 4{\mathrm{x}}^2 + 16\mathrm{x} - 9 = 0 \\ 4{\mathrm{x}}^2 + 18\mathrm{x} - 2\mathrm{x}- 9 = 0 \\ 2\mathrm{x} ( 2\mathrm{x} + 9 ) - 1 ( 2\mathrm{x} + 9 ) = 0 \\ ( 2\mathrm{x} + 9 ) ( 2\mathrm{x} - 1 ) = 0 \\ \mathrm{x} = -\frac{9}{2}, \frac{1}{2} \\ \mathrm{For} \mathrm{x} = -\frac{9}{2}, {\mathrm{y}}^2 = 4 \left(-\frac{9}{2}\right), \mathrm{which} \mathrm{is} \mathrm{not} \mathrm{possible}, \mathrm{hence} \mathrm{x} = \frac{1}{2} \\ \mathrm{Therefore}, \mathrm{the} \mathrm{given} \mathrm{curves} \mathrm{intersect} \mathrm{at} \mathrm{x} = \frac{1}{2}. \end{gathered}$$

Required area of the region bound by the two curves

$$\begin{gathered} = 2 {\int }_0^{\frac{1}{2}} 2\sqrt{\mathrm{x}}\mathrm{dx} + 2 {\int }_{\frac{1}{2}}^{\frac{3}{2}} \sqrt{\frac{9}{4} - {\mathrm{x}}^2} \mathrm{dx} \\ = 4 {\left[\frac{2}{3} {\mathrm{x}}^{\frac{3}{2}}\right]}_0^{\frac{1}{2}} + 2 {\left[\frac{\mathrm{x}}{2} \sqrt{\frac{9}{4} - {\mathrm{x}}^2} + \frac{9}{8} {\sin }^{-1} \left(\frac{2\mathrm{x}}{3}\right)\right]}_{\frac{1}{2}}^{\frac{3}{2}} \\ = \frac{8}{3}{\left(\frac{1}{8}\right)}^{\frac{1}{2}} + 2 \left[ 0 +\frac{9}{8} {\sin }^{-1} 1 - \frac{1}{4} \sqrt{2} - \frac{9}{8} {\sin }^{-1} \left(\frac{1}{3}\right)\right] \\ = \frac{8}{3}\left(\frac{1}{2\sqrt{2}}\right) + \frac{9}{4} \left(\frac{\mathrm{\pi }}{2}\right) - \frac{\sqrt{2}}{2} - \frac{9}{4} {\sin }^{-1} \left(\frac{1}{3}\right) \\ = \frac{2\sqrt{2}}{3} + \frac{9\mathrm{\pi }}{8} - \frac{\sqrt{2}}{2} -\frac{9}{4} {\sin }^{-1} \left(\frac{1}{3}\right) \end{gathered}$$

The point of intersection of the 

Parabolas  y² = 4x   and   x² = 4y are ( 0, 0 ) and ( 4, 4 )

Now the area of the region OAQBO bounded by curves  y² = 4x  and  x² = 4y, 

$$\begin{gathered} {\int }_0^4\left( 2 \sqrt{\mathrm{x}} - \frac{{\mathrm{x}}^2}{4}\right) \mathrm{dx} = {\left[ 2 \frac{{\mathrm{x}}^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}} - \frac{{\mathrm{x}}^3}{12}\right]}_0^4 = \frac{32}{3} - \frac{16}{3} = \frac{16}{3}\mathrm{sq}. \mathrm{units} \\ ..............\left(\mathrm{i}\right) \end{gathered}$$

Again, the area of the region OPQAO bounded by the curve  x² = 4y ,  x = 0,

x = 4  and  the  x - axis,

${\int }_0^4 \frac{{\mathrm{x}}^2}{4} \mathrm{dx} = {\left[\frac{{\mathrm{x}}^3}{12}\right]}_0^4 = \left(\frac{64}{12}\right) = \frac{16}{3} \mathrm{sq}. \mathrm{units} ..............\left(\mathrm{ii}\right)$

Similarly, the area of the region OBQRO bounded by the curve y² = 4x, the y-axis, y = 0  and  y = 4

${\int }_0^4 \frac{{\mathrm{y}}^2}{4} \mathrm{dy} = {\left[\frac{{\mathrm{y}}^3}{12}\right]}_0^4 = \frac{16}{3} \mathrm{sq}. \mathrm{units} ............\left(\mathrm{iii}\right)$

From (i), (ii), and (iii) it is concluded that the area of the region OAQBO =

area of the region OPQAO = area of the region OBQRO, i. e., area bounded

by parabolas  y² = 4x  and  x² = 4y  divides the area of the square into

three equal parts.

Given ellipse

$$\begin{gathered} \Rightarrow \frac{{\mathrm{x}}^2}{9} + \frac{{\mathrm{y}}^2}{4} = 1 \\ \Rightarrow \mathrm{y} = \frac{2}{3} \sqrt{ 9 - {\mathrm{x}}^2} \\ \mathrm{Given} \mathrm{line} \frac{\mathrm{x}}{3} + \frac{\mathrm{y}}{2} = 1 \\ \Rightarrow \mathrm{y} = \left( 2 - \frac{2 \mathrm{x}}{3}\right) \\ \mathrm{Required} \mathrm{Area} \left\{ \left( \mathrm{x}, \mathrm{y} \right): \frac{{\mathrm{x}}^2}{9} + \frac{{\mathrm{y}}^2}{4} \le 1 \le \frac{\mathrm{x}}{3} + \frac{\mathrm{y}}{2} \right\} \mathrm{is} \mathrm{given} \mathrm{below} \end{gathered}$$

$$\begin{gathered} \mathrm{Required} \mathrm{Area} = {\int }_0^3 \left( {\mathrm{y}}_1 - {\mathrm{y}}_2 \right) \mathrm{dx} \\ = {\int }_0^3 \left[\frac{2}{3} \sqrt{ 9 - {\mathrm{x}}^2} - \left( 2 - \frac{2\mathrm{x}}{3}\right) \right] \mathrm{dx} \\ = {\left[ \frac{2}{3} \left( \frac{\mathrm{x}}{2} \sqrt{ 9 - {\mathrm{x}}^2} + \frac{9}{2} {\sin }^{-1} \frac{\mathrm{x}}{3} \right) - 2\mathrm{x} + \frac{{\mathrm{x}}^2}{3}\right]}_0^3 \\ = \left[ \frac{2}{3} \left( \frac{9}{2} {\sin }^{-1} 1 \right) - 6 + 3 \right] - 0 \\ = 3 \times \frac{\mathrm{\pi }}{2} - 3 = \frac{3}{2} \left( \mathrm{\pi } - 2 \right) \mathrm{sq}. \mathrm{units} \end{gathered}$$

Equations of the lines are   y = 2 x + 1,    y = 3 x + 1     and   x = 4.

Let  y₁ = 2 x + 1,   y₂ = 3 x + 1 

Now area of the triangle bounded by the given lines,

$$\begin{gathered} = {}_0{}^4\left( {\mathrm{y}}_2 - {\mathrm{y}}_1 \right) \mathrm{dx} \\ = {}_0{}^4\left[\left( 3 \mathrm{x} + 1 \right) - \left( 2 \mathrm{x} + 1 \right) \right] \mathrm{dx} \\ = {}_0{}^4\mathrm{x} \mathrm{dx} \\ = \frac{1}{2} {\left[ {\mathrm{x}}^2 \right]}_0^4 \\ = \frac{1}{2} \left( 4^2 - 0^2 \right) \\ = \frac{1}{2} \times 16 \end{gathered}$$

= 8 sq. units

Thus, the area of the required triangular region is  8 square units

C.

Statement-1 is true, Statement-2 is false

∴ AD : DB = 2√2:√5
∴ OD is angle bisector of angle AOB
∴ Statement: 1 true
Statement : 2 false

C.

9

Equation of tangent at (2, 3) 

≡ x – 2y + 4 = 0
Required Area

D.

4/3

Solving the equations we get the points of intersection (–2, 1) and (–2,  1)The bounded region is shown as shaded region

B.

π

A.

C.

1/6

A.

1

D.

1 : 1 : 1

y² = 4x and x² = 4y are symmetric about line y = x

C.

A.

1

B.

$\frac{1}{2}\left(\sqrt{3-1}\right)$

18x2-9πx + π² = 0

(6x -π)(3x-π) = 0

$$\begin{gathered} \therefore x = \frac{\pi }{6},\frac{\pi }{3} \\ \alpha = \frac{\pi }{6}, \beta = \frac{\pi }{3} \\ y = \left(gof\right)\left(x\right) =\cos x \\ Area = {\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}} \cos x dx = (\sin x {)}_{\frac{\pi }{6}}^{\frac{\pi }{3}} \\ = \frac{\sqrt{3}}{2} -\frac{1}{2} \\ = \frac{1}{2}(\sqrt{3}-1) sq.units \end{gathered}$$

C.

13/9

$$\begin{gathered} 8 \cos x.\left({\cos }^2 \frac{\pi }{6}-{\sin }^2 x - \frac{1}{2}\right) =1 \\ \Rightarrow 8 \cos x\left(\frac{3}{4}-\frac{1}{2}-1 + {\cos }^2 x\right) = 1 \\ \Rightarrow 8 \cos x\left(\frac{-3+4 {\cos }^2 x}{4}\right) = 1 \\ \Rightarrow \cos 3x = 1 \\ \Rightarrow \cos 3x = \frac{1}{2} \\ \Rightarrow 3x = \frac{\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3} \\ \Rightarrow x = \frac{\pi }{9},\frac{5\pi }{9},\frac{7\pi }{9} \\ \Rightarrow sum = \frac{13\pi }{9} \\ \Rightarrow k = \frac{13}{9} \end{gathered}$$