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Question paper 1
1. (i) If a binary operation is defined by a'* b: ab then 2 * 2 is equal to:
(a) 4 (b) 2 (c) 9 (d) 8
Answer :
(a) 4
(ii) sin-1(1) is equal to :
(a) o (b) \({n\over 6}\) (c) \({n\over 2}\) (d) \({n\over 3}\)
Answer :
(c) \({n\over 2}\)
(iii) If order of matrix A is 4 * 3 and order of matrix B is 3 * 5 then order of matrix B'A' is:
a) 5 * 3 b) 4*5 c) 5 * 4 d) 3 * 2
Answer :
c) 5 * 4
(vi) If y: cos x then at X = \({n\over 2}\) , y2 is equal to:
a) -1 b) 1 c) 0 d) 1/2
Answer :
(c) 0
(v) \( \int\limits_0^{x/2}\) \({sin^{3/2} x \ \over sin^{3/2} x + cos^{3/2}x}\) dx is equal to :
a) 0 b) \({\pi \over 2} \) c) \({\pi \over 3}\) d) \({\pi \over 4} \)
Answer :
(d) \({\pi \over 4} \)
(vi) If .\({ \sqrt{3}}\) a .b = lA * Bl then angle between, vector a and vector b is :
a) \({\pi \over 2} \) b) \({\pi \over 6 } \) c) \({\pi \over 3}\) d) \({\pi \over 4} \)
Answer :
(b) \({\pi \over 6 } \)
(vii) Direction ratio of line given by \({ x-1\over 3} = { 6-2y\over 10} = { 1-z\over -7} \) are:
(a) < 3, 10, -7 > (c) < 3, -5, 7>
(b) < 3, 5, 7> (d) < 3, 5, -7>
Answer :
(d) < 3, 5, -7>
(viii) If P(A) = \({1 \over 2}\), P(B)= \({3 \over 8}\) and P(A \(\bigcap\) B) = \({1 \over 5}\) then P(B l A) is equal to:
a) \({ 2\over5} \) b) \({8\over 15 } \) c) \({2 \over 3}\) d) \({5\over 8} \)
Answer:
b) \({8\over 15 } \)
2. If A = \(\begin{bmatrix} 2 \\[0.3em] -4 \\[0.3em] 1 \end{bmatrix}\), B = [ 5 3 -1 ] then verify that (AB)' = B'A'
Answer:
A = \(\begin{bmatrix} 2 \\[0.3em] -4 \\[0.3em] 1 \end{bmatrix}\), B = [ 5 3 -1 ]
AB = \(\begin{bmatrix} 10 & 6 & -2 \\[0.3em] -20 & -12 &4 \\[0.3em] 5 & 3 & -1 \end{bmatrix}\)
\((AB)^{-1}\) = \(\begin{bmatrix} 10 & -20 & 5 \\[0.3em] 6 & -12 &3 \\[0.3em] -2 & 4 & -1 \end{bmatrix}\)
\(B^1 A^1\) = \(\begin{bmatrix} 5 \\[0.3em] 3 \\[0.3em] -1 \end{bmatrix}\)[ 2 -4 1 ]
\(B^1 A^1\) = \(\begin{bmatrix} 10 & -20 & 5 \\[0.3em] 6 & -12 &3 \\[0.3em] -2 & 4 & -1 \end{bmatrix}\)
\((AB)^{-1}\) = \(B^1 A^1\)
3) If Y = \({{sin^-1} }\) ( \({2x\over 1+x^2}\) ) then find \({dy\over dx}\)
Answer:
Y = \({{sin^-1} }\) ( \({2x\over 1+x^2}\) )
put x = tanQ
y = \({sin^{-1}} ({ 2tanQ \over 1+ tan^2Q})\)
y = \(sin^{-1} sin 2Q\)
y = 2Q = \(2 tan^{-1}Q\)
\({dy \over dx} = {2 \over 1+ x^2}\)
4) Evaluate \({\int {sin^4x cos^3x} }\) dx
Answer:
I = \({\int {sin^4x cos^3x} }\) dx
= \({\int {sin^4x cos^2x cosx} }\) dx
= \({\int {sin^4x } (1 - sin ^2x) cosx }\) dx
put sin x = y
cosx dx = dy
I = \({\int y^4{(1 - y^2)} dy}\)
= \({\int y^4 dy} - {\int y^6 dy } \)
= \({y^5 \over 5} - {y^7 \over 7 }+ c = {sin ^5 x\over 5}- {sin^7x\over7 } +c\)
5) Evaluate \({\int {dx \over {x^2-4x + 13}} }\)
Answer:
\({\int {dx \over {x^2-4x + 13}} }\)
I = \({\int {dx \over {(x-2)^2 - 4 + 13}} }\)
Put x = 2 = t
dx = dt
I = \({\int {dt \over {t^2 - (3)^2}}} \)
= \({1 \over 3 } tan^{-1} {t \over 3} + c\)
I = \({1 \over 3 } tan^{-1} ({x-2 \over 3}) + c\)
6. Find Particular solution of differential equation cos ( \({dy \over dx}\) ) = \({1 \over 5}\) , y (0) = 2
Answer:
5dy = 1dx
5y = x+c
put y(0) = 2
5(2) = 0 + c
c = 10
5y = x + 10
7. Find the integrating factor for the differential equation cot x \({ dy \over dx}\) + y = 2x + \({x^2}\)
Answer:
cot x \({ dy \over dx}\) + y = 2x + \({x^2}\)
\({ dy \over dx}\) + \({y \over x} = 2 + x\)
I.F = \(e^{\int pdx}\)
I.F = \(e^{\int {1 \over x}dx} = e^{log x} = x\)
8) Find the angle between plane 3x + 4y - z = 8 and line \({x-1\over 2}\) = \({2-y\over 7}\) = \({3z + 6\over 12}\)
Answer:
The equation of the line
\({x-1\over 2}\) = \({2-y\over 7}\) = \({3z + 6\over 12}\)
\({x-1\over 2}\) = \({2-y\over 7}\) = \(3({z + 6\over 4})\)
The Direction Ratio are : 2, -7, 4
The Equation of the plane is :
3x + 4y - z = 8
ie : 3x + 4y - z - 8 = 0
Direction Ratio of the normal to the plane are 3, 4, -1
Let Q be the angle between line & the plane
Since = \({(3)(2) + 4(7) + 4(-1) \over \sqrt{4+ 49 + 16 } \sqrt{9 + 16 + 1}}\)
= \(6 + -28 - 4 \over \sqrt{69} \sqrt {26}\)
= \(2 - 28 \over \sqrt{69} \sqrt {26}\)
= \(- 26 \over \sqrt{69} \sqrt {26}\)
9) Probability of A,B and C of solving the problem are 1/3 , 1/2 and 1/4 respectively . If they all try to solve the problem then find the probability that exactly one of them will solve the problem
Answer:
Given that
P (\(\bar{A}\)) = \(1 \over 3 \) P (B) = \(1 \over 2\) P (C) = \(1 \over 4 \)
P(A) = 1 - P (A) = 1 - \(1 \over 3 \) = \(2 \over 3 \)
P(\(\bar{B}\)) = 1 - P (B) = 1 - \(1 \over 2 \) = \(1 \over 2 \)
P(\(\bar{C}\)) = 1 - \(1 \over 4 \) = \(3 \over 4 \)
Req Prob
= P(A) P(\(\bar{B}\)) P(\(\bar{C}\)) + P(\(\bar{A}\)) P(B) P(\(\bar{C}\)) + P(\(\bar{A}\)) P(\(\bar{B}\)) P(C)
= \(({1 \over 3})({1 \over 2})({3 \over 4})+({2 \over 3})({1\over2})({3 \over 4})+({2 \over 3})({1\over2})({1 \over 4})\)
= \(11 \over 24\)
10) Show that function f: R \(\rightarrow\) R, f(x) = \({2x+5\over 8}\) is invertible. Also find inverse of f.
Answer:
f(x) = \({2x+5\over 8}\) , x \(\leftarrow\) R
7(x1) = f(x2)
x1x2 \(\leftarrow\) R
\({2x_1 + 5 \over 8} = {2x_2 + 5 \over 8}\)
=> x1 = x2 f is 1 - 1
k = \({2x + 5 \over 8} \)
=> 8k = 2x+ 5
=> 8k - 5 = 2x
=> x = \({8k - 5 \over 2} \)
=> f is onto => f is invertible
Ley k \(\leftarrow\) R
f(x) = k
f(k) = x
\({2k + 5 \over 8} \) = x
2k = 8x - 5
k = \({8x + 5 \over 2} \)
for x \(\leftarrow\) R
f(x) = \({8x + 5 \over 2} \)
11) Show that \({tan^{-1} 1\over 3} + {tan^{-1} 1\over 5} = {1\over 2} {cos^{-1} 33\over 65}\)
Answer:
\({tan^{-1} 1\over 3} + {tan^{-1} 1\over 5} = {1\over 2} {cos^{-1} 33\over 65}\)
= \( {tan^{-1}} ({1 \over 3}) + {tan^{-1}} ({1 \over 5})\)
= \({tan^{-1}} [{{1 \over 3}+{1 \over 5} \over 1 - {1\over 3}+ {1 \over 5}}]\)
= \({tan^{-1}} [{{8 \over 15} \over 1 - {1\over 15}}]\)
= \({tan^{-1}} [{{8 \over 15} \over {14\over 15}}]\)
= \({tan^{-1}} [{{8 \over 14} }] = {tan^{-1}} [{{4 \over 7} }]\)
Put \({1 \over 2 }cos^{-1} ({33 \over 65}) = Q\)
= \(cos^{-1} ({33 \over 65}) = 2Q\)
\(cos 2Q = {33 \over 65}\)
\(1 + tan^2Q \over 1 + tan^2Q\) = \(33 \over 65\)
\(65 - 65 tan^2 Q = 33 + 33 tan^2Q \)
\((-65 - 33) tan^2 Q = 65 + 33 \)
\(- 98 tan^2 Q = -32 \)
\( tan^2 Q = {32 \over 98}\)
\( tan^2 Q = {16 \over 49}\)
\( tanQ = {4\over 7}\)
\(Q = tan^{-1} ({4\over 7})\)
12) Express \(\begin{bmatrix} 6 & -4 & 5 \\[0.3em] 1 & 4 & -2\\[0.3em] 7 & 5 & 9 \end{bmatrix}\)as sum of symmetric matrix and a skew-symmetric matrix,
Answer:
A = \(\begin{bmatrix} 6 & -4 & 5 \\[0.3em] 1 & 4 & -2\\[0.3em] 7 & 5 & 9 \end{bmatrix}\)
AT = \(\begin{bmatrix} 6 & 1 & 7 \\[0.3em] -4 & 4 & 5\\[0.3em] 5 & -2 & 9 \end{bmatrix}\)
For symmetric matrix = \(A + A^T \over 2\)
\(A + A^T \over 2\) = \(1 \over 2\) \(\begin{bmatrix} 6 & -4 & 5 \\[0.3em] 1 & 4 & -2\\[0.3em] 7 & 5 & 9 \end{bmatrix}\) + \(\begin{bmatrix} 6 & 1 & 7 \\[0.3em] -4 & 4 & 5\\[0.3em] 5 & -2 & 9 \end{bmatrix}\)
= \(1 \over 2\) \(\begin{bmatrix} 12 & -3 & 12 \\[0.3em] -3 & 8 & 3\\[0.3em] 12 & 3 & 18 \end{bmatrix}\)
For skew-symmetric matrix = \(A - A^T \over 2\)
\(A - A^T \over 2\) = \(1 \over 2\) \(\begin{bmatrix} 6 & -4 & 5 \\[0.3em] 1 & 4 & -2\\[0.3em] 7 & 5 & 9 \end{bmatrix}\) + \(\begin{bmatrix} 6 & 1 & 7 \\[0.3em] -4 & 4 & 5\\[0.3em] 5 & -2 & 9 \end{bmatrix}\)
= \(\begin{bmatrix} 0 & -5 & -2 \\[0.3em] 5 & 0 & -7\\[0.3em] 2 & 7 & 0 \end{bmatrix}\)
13. Using differentials find approximate value of \(\sqrt 360\)
Answer:
Take y = \(\sqrt x\) , x = 361
dx = 8x = -1 so that x - 8x = 360
Now y + 8y = \(\sqrt{x + 3x}\)
y + 8y - y = \(\sqrt{x + 8x}\) - \(\sqrt x\)
8y = \(\sqrt{360} \) - \(\sqrt{361} \)
\(\sqrt{360} = 8y + 19\)
Now 3y is opp eqaul to dy and dy = \(dy\over dx\) = \(1 \over 2 \sqrt x\)dx
dy = \(1 \over 2 * 19 \) dx = -\(1 \over38 \)
\(\sqrt{360} \) = -\(1 \over38 \) + 19
\(\sqrt{360} \) = - 0.026 + 19
\(\sqrt{360} \) = 18.974
14. Evaluate \(\int\limits_1^3\) \((x^2+4) dx \) as limit of a sum
Answer:
15. Using integration find the area of triangle whose sides are given by equations y= x + 1, y = 3x + 1, x=5
Answer:
y= x + 1, --------(i)
y = 3x + 1, ----------(ii)
x=5 ----------(iii)
Subtrating (i) from (ii)
x = 0
putting x = 0 in (i) y = 1
Line 1 and 2 intersect in A(0,1)
From (i) & (iii) we get x = 5 , y = 5+ 1 = 6
Line 1 and 3 intersect in B(5, 6)
From (ii) and (iii)
x = 5 , y = 16
Line 2 and 3 intersect in C(5 , 16)
vertices of the ABC are A (0, 1); B(5, 6); C(5, 16);
Req Area = Area of triangle ABC
= Area of triangle OMC - Area of triangle OMB
= \(\int_0^5 (3x + 1)dx - \int_0^5 (x + 1)dx\)
= \({3x^2 \over 2} + x |_0^5 - {x^2 \over 2} + x |_0^5\)
= \({3x^2 - x^2\over 2} |_0^5\)
= 24 sq units
16. Find particular solution of differential equation \(x^2dy -(3x^2+xy+ y^2)dx = 0, y(1) = 1\)
Answer:
\(x^2dy -(3x^2+xy+ y^2)dx = 0\)
\(x^2dy = (3x^2+xy+ y^2)dx \)
\(({dy \over dx}) = {3x^2 + xy + y^2 \over x^2}\)
\(({dy \over dx}) = 3 +{ xy + y^2 \over x^2}\)
\({dy \over dx} = { xy + y^2 \over x^2} + 3\)
Put y = Vx & \({dy \over dx } = x {dv \over dx} + v\)
\( x {dv \over dx} + v\) = \( { x(Vx) + (Vx)^2 \over x^2} + 3\)
\( x {dv \over dx} + v\) = \({ x^2 (V + V^2)\over x^2} + 3\)
\( x {dv \over dx}\) = \(V^2 + 3\)
\({dv \over V^2+ 3} = {dx \over x}\)
\({1 \over \sqrt 3} tan^{-1} {V \over \sqrt3} = log x + c\)
\({1 \over \sqrt 3} tan^{-1} {V \over \sqrt3x} = log x + c\)
\({1 \over \sqrt 3} tan^{-1} ({1 \over \sqrt3}) = c\)
\({1 \over \sqrt 3} { \pi \over 6} = c\)
17) Adjacent sides of a parallelogram are given by the vectors \(2\hat{i} - \hat{j} + 2\hat{k}\) and \(\hat{i} + 5\hat{j} - \hat{k}\) find a unit vector in a drection of it's diagonal. Also find the area of parellelogram.
Answer:
Let ABCD be the parallelogram with
\(\overrightarrow{AB}\) = \(2\hat{i} - \hat{j} + 2\hat{k}\)
\(\overrightarrow{AD}\) = \(\hat{i} + 5\hat{j} - \hat{k}\)
Now = \(\overrightarrow{AB}\) = \(\overrightarrow{AB}\) + \(\overrightarrow{BC}\)
= \(\overrightarrow{AB}\) + \(\overrightarrow{AD}\)
\(\overrightarrow{AC}\) = (\(2\hat{i} - \hat{j} + 2\hat{k}\)) + (\(\hat{i} + 5\hat{j} - \hat{k}\))
\(\overrightarrow{AC}\) = (\(3\hat{i} + 4\hat{j} + \hat{k}\))
Hence , a limit vector in the direction
\(\overrightarrow{AC}\) = \({1 \over |\overrightarrow{AC} |} (\overrightarrow{AC})\)
= \({3\hat{i} + 4\hat{j} + \hat{k}} \over \sqrt {9 + 16 + 1}\)
= \(1 \over \sqrt {26}\)\((3\hat{i} + 4\hat{j} + \hat{k})\)
18) BagI contains 2 black and 8 red balls, bagII contains 7 black and 3 red balls and bag III contains 5 black and 5 red balls. One bag is chosen at random and a ball is drawn from it which is found to be red. Find the probability that the ball is drawn from bag II.
or
Two cards are drawn (without replacemant) from a well shuffled deck of 52 cards. Find probability distribution and mean of number of cards numbered 4.
Answer:
19. Solve the following sysJeur oflinear equations by matrix method :
3 x+y +z=10, 2x-y-z=0, x-y + 2z = 1
or
Using elementary transformation find the inverse of \(\begin{bmatrix} 3 & 2 & 1 \\[0.3em] 2 & 4 & 3 \\[0.3em] 2 & -1 & 2 \end{bmatrix}\)
Answer:
3 x+y +z=10,
2x-y-z=0,
x-y + 2z = 1
Let A = \(\begin{bmatrix} 3 & 1 & 1 \\[0.3em] 2 & -1 & -1 \\[0.3em] 1 & -1 & 2 \end{bmatrix}\); x = \(\begin{bmatrix} x \\[0.3em] y \\[0.3em] z \end{bmatrix}\)
and B = \(\begin{bmatrix} 10 \\[0.3em] 0 \\[0.3em] 1 \end{bmatrix}\)
|A| = 3(-2 -1) -1 (4 + 1) +1 (-2 + 1)
|A| = 3(-3) - 5 - 1 = - 9 - 6 = - 15 != 0
\(A_{11}\) = \(\begin{vmatrix} -1 & -1 \\[0.3em] -1 & 2 \\[0.3em] \end{vmatrix}\) = - 2 - 1 = -3
\(A_{12}\) = -\(\begin{vmatrix} 2 & -1 \\[0.3em] 1 & 2 \\[0.3em] \end{vmatrix}\) = - (4 + 1) = - 5
\(A_{12}\) = \(\begin{vmatrix} 2 & -1 \\[0.3em] 1 & -1 \\[0.3em] \end{vmatrix}\) = - 2 + 1 = - 1
\(A_{21}\) = \(\begin{vmatrix} 1 & 1 \\[0.3em] -1 & 2 \\[0.3em] \end{vmatrix}\) = - (2 + 1) = - 3
\(A_{22}\) = \(\begin{vmatrix} 3 & 1 \\[0.3em] 1 & 2 \\[0.3em] \end{vmatrix}\) = 6 - 1 = 5
\(A_{23}\) = - \(\begin{vmatrix} 3 & 1 \\[0.3em] 1 & -1 \\[0.3em] \end{vmatrix}\) = - (-3 - 1) = 4
\(A_{31}\) = \(\begin{vmatrix} 1 & 1 \\[0.3em] -1 & -1 \\[0.3em] \end{vmatrix}\) = - 1 + 1 = 0
\(A_{32} \) = - \(\begin{vmatrix} 3 & 1 \\[0.3em] 2 & -1 \\[0.3em] \end{vmatrix}\) = - (- 3 - 2) = 5
\(A_{33}\) = \(\begin{vmatrix} 3 & 1 \\[0.3em] 2 & -1 \\[0.3em] \end{vmatrix}\) = - 3 - 2 = - 5
adj A = \(\begin{bmatrix} -3 & -5 & -1 \\[0.3em] -3 & 5 & 4 \\[0.3em] 0 & 5 & -5 \end{bmatrix}^t\) = \(\begin{bmatrix} 3 & -3 & 0 \\[0.3em] -5 & 5 & 5 \\[0.3em] -1 & 4 & -5 \end{bmatrix}\)
\(A^{-1}\) = \(adj A \over |A|\) = -\(1 \over 15\)\(\begin{bmatrix} 3 & -3 & 0 \\[0.3em] -5 & 5 & 5 \\[0.3em] -1 & 4 & -5 \end{bmatrix}\)
x = \(A^{-1} B\)
x = -\(1 \over 15\)\(\begin{bmatrix} 3 & -3 & 0 \\[0.3em] -5 & 5 & 5 \\[0.3em] -1 & 4 & -5 \end{bmatrix}\)\(\begin{bmatrix} 10 \\[0.3em] 0 \\[0.3em] 1 \end{bmatrix}\)
= -\(1 \over 15\) \(\begin{bmatrix} 30 + 0 + 0 \\[0.3em] -50 + 0 + 5 \\[0.3em] -10 + 0 - 5 \end{bmatrix}\)
= \(1 \over 15\)\(\begin{bmatrix} 30 \\[0.3em] -45 \\[0.3em] -15 \end{bmatrix}\)= \(\begin{bmatrix} -2 \\[0.3em] 3 \\[0.3em] 1 \end{bmatrix}\)
x = -2 ; y = 3 ; z = 1
20) A wire of length 25 cm is to be cut off into two pieces. One piece is to be made into a circle and other into a square. What should be the lengths of'two pieces so that combined area of,circle and square is minimum ?
Answer:
21) Find the image of the point (5, - 3, l) in the plane 2x - 2y - 3z = 10
Answer:
The equation of the plane is 2x - 2y - 3z = 10 -------------(i)
From P (5,3,-1) draw PM plane and perpendicular is to P' s.t . M is the mid-point of PP' Then P' (\(\alpha \beta \gamma\)) is Image of P
Divertion ratio of PM are 2, -2 , -3
The equation of PM are:
\(x- 5 \over 2\) = \(y + 3 \over -2\) = \(z - 3 \over 1\)
Any pt on it is
M(\(\alpha + 5 , -\beta -3, \gamma - 3\))
M lies on (i)
2(r + 5) - 2(-r -3) - 3( r - 3 ) = 10
2r + 10 + 2r + 6 -3r + 9 = 10
r = -15
M is (-10, 12, -18)
since M is mid point of PP'
\({\alpha + 5 \over 2 }= -10 {\beta - 3 \over 2 } = 12\)
\(\alpha\)= -20 - 5 \(\beta \) = 24 + 3
\(\alpha\) = -25 \(\beta \) = 27
\(\gamma + 1 \over 2\) = -18
\(\gamma\) = -37
P' (-25, 27, -37) which is the image of P in the plane
Or
Find the shortest distance between the lines :
\({x+1 \over 4}= {y-3 \over -6} = {z+1 \over 1} and {x+3 \over 3}={y-5 \over 2}={z-7 \over 6}\)
Answer :
The equation of two line :
\({x+1 \over 4}= {y-3 \over -6} = {z+1 \over 1} \) or \({x-(-1) \over 4}= {y-3 \over -6} = {z-(-1) \over 1} \)
And
\( {x+3 \over 3}={y-5 \over 2}={z-7 \over 6}\) or \( {x- (-3) \over 3}={y-5 \over 2}={z-7 \over 6}\)
\(x_1 = -1 , y_1 = -3, z_1 , -1; \)
\(a_1 = 4 , b_1 = -6 , c_1 = 1\)
and
\(x_2 = -3 , y_2 = 5, z_2 = 7; \)
\(a_2 = 5 , b_1 = -2 , c_1 = 6\)
\(x_2 - x_1 = -3 - (-1 ) = -3 + 1 = -2\)
\(y_2 - y_1 = 5 + 3 = 8\)
\(z_2 - z_1 = 7 + 1 = 8\)
\(\begin{vmatrix} x_2- x_1 & y_2 -y_1 & z_2 - z_1 \\[0.3em] a_1 & b_1 & c_1 \\[0.3em] a_2 & b_2 & c_2 \end{vmatrix}\)
= \(\begin{vmatrix} -2 & 8 & 8 \\[0.3em] 4 & -6 & 1 \\[0.3em] 5 & 2 & 6 \end{vmatrix}\)
= \(-2(-36 -2) -8 (24 -5) +8(8 + 30)\)
= \(-2(-38) -8 (19) +8(38)\)
= 76 - 152 + 304
= 228
\(\sqrt {(b_1 c_2 - b_2 c_1)^2+ (c_1 a_2 - c_2 a_1)^2 + (a_1 b_2 - a_2 b_1)^2}\)
= \(\sqrt {((-6) (6) - (2) (11))^2+ ((1) (5) - (6) (4))^2 + ((4) (2) - (5) (-6))^2}\)
= \(\sqrt {(36 - 2)^2+ (5 -24)^2 + (8 + 30)^2}\)
= \(\sqrt {(36 - 2)^2+ (5 -24)^2 + (8 + 30)^2}\)
= \(\sqrt {(34)^2+ (-19)^2 + (38)^2}\)
= \(\sqrt {(1156 + 361 + 1444)}\)
= \(\sqrt {2961}\)
S.D = \(|{228 \over \sqrt {2961}}|\)
= \({228 \over 3\sqrt {329}}\)
S.D = \({76 \over \sqrt {329}}\)
22) Maximize Z = 12x + 24y subject to the constrains x+y >= 5, 5x+7y <= 35, x-y>=0, x,y, >-0 graphically.
Answer:
Or
One kind of cake requires 300 gm of flour and 15 gm of fat and another kind of cakerequires 150 gm of flour and 30 gm of fat. Find the maximum number of cakes that can be made from 7. 5 kg of flour and 600 gm of fat, From a linear programing problem and solve it graphically
Answer:
Let x and y be the number of cakes of first and second type that can be made. Clearly x>= 0 , y >= 0
Let Z be the no of cakes
Kind Number of cakes Flour required
(in grams)
Fat required
(in grams)
I
II
x
y
300x
150y
15x
30y
Total x+y 300x + 150y 15x + 30y Mathematical formulation of the L.P.P is as follow:
Maximise : Z = x + y
subject to the constrains
300x + 150y <= 7500 i.e 2x + y <= 50
15x + 30y <= 600 i.e x + 2y <= 40
x,y >= 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfied x>= 0 and y >= 0 lies in the first quadrant
Now we draw the graph of 2x + y = 50
For x = 0 , y = 50
For y = 0 , 2x = 50 or x = 25;
lines meet OX in A(25,0) and OY in L(0,50)
Again we draw the graph of x + 2y = 40
For x = 0 , 2y = 40 or y = 20
For y = 0 , x = 40
Lines meet OX in B(40, 0) and OY in M(0, 20)
Since feasible region satisfied of all the constraints
OACM is the feasible region
The corner point are O(0,0), A(25,0), C(20,10), M(0,20)
At O(0,0), Z = 0 + 0 = 0
A(25,0) Z = 25 + 0 = 25
C(20,10) Z = 20 + 10 = 30
M(0,20) Z = 0 + 20 = 20
Minimum value = 30 at (20,10)
Maximum number of cake is 20 of one kind and 10 of the second kind.