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1. a = – b ⇒ vectors have same magnitude, but opposite directions. Hence | a | = | b | is
true.
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Converse may not be true, e.g. a = it + tj – 2kt and b = 2it + tj + kt have same magnitude, i.e.
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| a | = | b | but a ! b or a ! –b.
dy
2.
Consider
dx
dy
= – a sin x
+ y tan x = – a sin x + a cos x. tan x
dx
Hence, solution
= – a sin x + a sin x = 0
3.
3 2 6
Direction cosines of perpendicular from origin are ± < , − , >
7 7 7
3 2
6
∴ Equation is ± ( x– y + z ) = 7 ⇒ ± (3x – 2y + 6z) = 49.
7 7
7
4. Given equation is c
dy 4
d2 y
m + 3y 2 = 0
dx
dx
Order = 2, degree = 1
5. Given vector a = it + tj + 2kt
Vector of magnitude
6 $ at =
6 units in the direction of a is
6⋅
it + tj + 2kt
1+1+4
= it + tj + 2kt
6. Let point (x, y) lies on the line joining the points (3, –5) and (0, 4). As points lie on the
line so, these points are collinear.
x
y 1
1
1
3 − 5 1 = 0 ⇒ [x(–9) – y(3) + 1(12)] = 0
2
2
0
4 1
⇒ –9x – 3y + 12 = 0 ⇒ 3x + y – 4 = 0
⇒
7. Number of students for thinking skills = x
Number of students for attitude towards school programmes = y
Number of students for participation in sports = z
According to the question
y + 2z = 4
⇒
0x + y + 2z = 4
⇒
x + 2z = 5
⇒
x + 0y + 2z = 5
⇒
x + 2y = 7
⇒
x + 2y + 0z = 7
1
Corresponding matrix equation is
0 1
0
1 2
>1
i.e.
For A–1:
AX = B, Its solution is X = A–1B
0 1
|A| = 1 0
1 2
Adj A =
∴
4
2 x
2 H > y H = >5 H
7
0 z
A–1 =
−4
> 4
2
...(i)
2
2 = 0 – 1(–2) + 2(2) = 6 ≠ 0
0
2l
−4
1H = > 2
2
−1
4
−2
1
−4
1
2
>
6
2
2
2H
−1
2
−2
2
Adj A
|A |
=
4
−2
1
2
2H
−1
Substituting in (i), we get
X=
⇒
−4
1
2
>
6
2
4
−2
1
2 4
− 16 + 20 + 14
1
2H>5 H = > 8 − 10 + 14 H
6
8+5 −7
−1 7
18
x
3
1
> yH = 6 >12H = >2H
6
z
1
⇒ x = 3, y = 2, z = 1
∴ Number of students awarded for thinking skills = 3
Number of students awarded for attitude towards school programme = 2
Number of students awarded for participation in sports = 1
Participation in physical activities and sports is of utmost importance as it inculcates
team spirit, fitness, decision making, sportsmanship.
8. Consider
⇒
⇒
⇒
⇒
⇒
2
sin–1 x + sin–1 2x =
π
3
π
– sin–1x
3
π
2x = sin c − sin − 1 xm
3
π
π
2x = sin cos(sin–1x) – cos sin (sin–1x)
3
3
1
3
⋅ 1 − x2 – ⋅ x
2x =
2
2
sin–1 2x =
3
5x
=
2
2
1 − x2
Squaring both sides, we get
25x2 = 3(1 – x2) ⇒ 28x2 = 3
⇒
x2 =
⇒
x =
3
⇒x=±
28
3
28
3
as x = −
28
3
does not satisfy the given equation.
28
OR
Let x = a tan θ ⇒ θ = tan–1 x
a
2
3
3a3 tan θ – a3 tan3 θ
3
a
x
–
x
–1 )
3
tan–1 )
=
tan
3
a (a2 – 3a2 tan2 θ)
a (a2 – 3x2)
= tan–1 )
a3 (3 tan θ – tan3 θ)
= tan–1 )
3 tan θ – tan3 θ
a3 (1 – 3 tan2 θ)
1 – 3 tan2 θ
x
= 3θ = 3tan–1
a
ax + 1, x # 3
9. Consider function f(x) = )
bx + 3, x > 3
For continuity at x = 3
...(i)
3
3 = tan–1(tan 3θ)
[from (i)]
Lim f(x) = Lim f(x) = f(3)
x → 3+
x → 3–
⇒
Lim f(3 – h) = Lim f(3 + h) = 3a + 1
h→0
h→0
⇒ Lim {a(3 – h) + 1} = Lim {b(3 + h) + 3} = 3a + 1
h→0
h→0
⇒
3a + 1 = 3b + 3 = 3a + 1
⇒
3a + 1 = 3b + 3 ⇒ 3a – 3b = 2
For values of a and b satisfying the relation 3a – 3b = 2, function is continuous at x = 3
10. Consider
xy = yx
Taking log on both sides, we get
ylogx = xlogy
Differentiating both sides w.r.t. x, we get
y$
⇒
⇒
1
1
+ log x $ yl = x $ $ yl + log y $ 1
y
x
y
x
yl= log x– G = log y –
y
x
y′ =
y (x log y – y)
x (y log x – x)
3
11. Consider curves
y2 = 4ax
and
...(i)
xy = c2
...(ii)
Differentiating (i) and (ii), with respect to x, we get
2y
dy
dy 2a
= 4a ⇒
=
dx
dx
y
...(iii)
dy
dy
y
+ y.1 = 0 ⇒
=–
dx
dx
x
If curves cut at right angles, then from (iii) and (iv), we get
and
x
...(iv)
y
2a
× c− m = –1 ⇒ x = 2a
x
y
From (i) we get
...(v)
y2 = 4a × 2a = 8a2
...(vi)
From (ii) squaring both sides, we get
x2y2 = c4
2
⇒
(2a) (8a2) = c4
⇒
32a4 = c4
12. Consider
I =
=
[from (v) and (vi)]
cos x
dx
y cos
3x
cos x
y
dx =
3
4 cos x − 3 cos x
y
1
4 cos2 x − 3
dx
Dividing numerator and denominator by cos2x, we get
I=
y
=
y
=
=
=
13. Consider
I=
Using property
4
sec2 x
4 − 3 sec2 x
sec2 x
1 − 3 tan2 x
1
y
1
3 1−t
2
dx =
y
sec2 x
4 − 3 − 3 tan2 x
dx
dx
Let
dt
⇒
3 tan x = t
3 sec2x dx = dt
1
1+t
log
H+ c
2
×
1
1
–t
3
1
>
1
2 3
y0
π
2
log
1 + 3 tan x
1 − 3 tan x
+c
x
dx
sin x + cos x
y0a f (x) dx = y0a f (a − x) dx , we get
...(i)
I=
y0
π
−x
2
π
2
π
π
sin c − x m + cos c − x m
2
2
π
π
−x
2
2
= y
dx
0 cos x + sin x
Adding (i) and (ii), we get
π
π
2
2
2I = y
dx
0 sin x + cos x
=
=
=
=
⇒
π
2
dx
...(ii)
π
y02 sin x 1+ cos x dx
π
2 2
π
2 2
π
2 2
y0
y0
y0
π
2
1
1
2
π
2
π
2
sin x +
1
π
sin c x + m
4
1
2
dx
cos x
dx
π
cosec c x + m dx
4
π
π
π
π 2
I=
= log cosec c x + m − cot c x + m G
4
4 0
4 2
=
=
=
=
=
π
π π
π π
π
π
; log cosec c + m − cot c + m − log cosec − cot E
2 4
2 4
4
4
4 2
π
4 2
π
4 2
π
4 2
π
2 2
; log sec
2 +1
log
2 −1
$ 2 log
log
π
π
+ tan − log
4
4
=
π
4 2
2 −1 E
^ 2 + 1h
2
log
2−1
2 +1
2 +1
14. General equation of a plane through the intersection of planes x + 3y + 6 = 0 and
3x – y – 4z = 0 is
x + 3y + 6 + λ(3x – y – 4z) = 0
⇒ (1 + 3λ)x + (3 – λ)y – 4λz + 6 = 0
...(i)
Solutions to CBSE Pariksha-2015 (Mathematics-XII) 5
If distance of (i) from origin is one unit, then
(1 + 3λ) × 0 + (3 – λ) × 0 – 4λ × 0 + 6
(1 + 3λ) 2 + (3 – λ) 2 + 16λ2
=1
Squaring both sides, we get
⇒
36 = 1 + 9λ2 + 6λ + 9 + λ2 – 6λ + 16λ2
26λ2 = 26 ⇒ λ2 = 1 ⇒ λ = ±1.
When λ = 1 from (i)
4x + 2y – 4z + 6 = 0, i.e. 2x + y – 2z + 3 = 0 is the required equation of the plane
When λ = – 1. from (i)
– 2x + 4y + 4z + 6 = 0, i.e. x – 2y – 2z – 3 = 0 is the required equation of the plane
OR
Given line is 3x = 2y = – z
y
x
z
= =
2
3 −6
Line parallel to the given line and passing
or
A (1, –2, 3)
B
through A(1, –2, 3) is
y+2
x−1
z−3
=
=
=λ
2
3
−6
General point on the line is B(2λ + 1, 3λ – 2, – 6λ + 3)
If this point lies on the plane x – y + z = 5
1
7
3
–6
9 11 15
2
Point of intersection is B c + 1, − 2,
+ 3m , i.e. B c , – ,
m
7
7
7
7
7 7
then 2λ + 1 – 3λ + 2 – 6λ + 3 = 5 ⇒ –7λ = – 1 ⇒ λ =
∴ Distance AB =
=
2
2
2
9
15
− 11
+ 2m + c
− 3m
c − 1m + c
7
7
7
9 36
4
=
+
+
49 49 49
49
= 1 unit
49
15. Given vectors are a = 2x2 it + 4x tj + kt and b = 7it – 2 tj + x kt
cos θ =
7 × 2 x 2 + 4 x × (− 2 ) + x × 1
4x 4 + 16x2 + 1 49 + 4 + x2
If angle between two vectors is obtuse then
14x2 – 8x + x < 0 ⇒ 14x2 – 7x < 0
⇒ 7 (2x – 1) < 0
6
3x = 2y = – z
x–y+z=5
⇒ x < 0, 2x – 1 > 0
or x > 0, 2x – 1 < 0
1
2
⇒ no solution
⇒ x < 0, x >
∴ For 0 < x <
or x > 0, x <
or 0 < x <
1
2
1
2
1
angle between vectors a and b is obtuse.
2
16. A : girl students
n(A) = 430
n(B) = number of girls studying in class XII = 10% of 430 = 43
P(B/A) =
43/1000
1
P (A + B)
=
=
430/1000 10
P (A)
OR
Given integers are 1 to 11.
There are 6 odd numbers and 5 even numbers.
E : sum of two integers choosen is even
A : both even;
B: both odd
5
6
C
C
P(E/A) = 11 2 ; P(E/B) = 11 2
C2
C2
Probability that both numbers are odd
6
=
=
−7
17. Let A = =
4
1
G
−1
We have A = IA ⇒ =
6
5
C2
6
C2 + C2
=
5
C2
6
C2
+
11
11
C2
C2
C2
15
15 3
=
=
10 + 15 25 5
1
1 0
−7
G==
GA
4 −1
0 1
[By performing R1 → R1 + 2R2]
=
1 −1
1
2
G==
GA
0
3
−4 − 7
[By performing R2 → R2 – 4R1]
=
1
2
1 −1
G==
GA
0
1
− 4/3 − 7/3
=
⇒
⇒
∴
P (E/A) + P (E/B)
=
C2
1 −1
1 2
G==
GA
4 −1
0 1
⇒
⇒
P^ E/Bh
11
1 0
− 1/3
G==
=
0 1
− 4/3
A–1 = =
− 1/3
− 4/3
− 1/3
GA
− 7/3
[By performing R2 →
1
R]
3 2
[By performing R1 → R1 + R2]
− 1/3
G
− 7/3
7
OR
Consider equations 2X + 3Y = =
3X + 2Y = =
and
2
4
−2
1
3
G
0
...(i)
2
G
−5
...(ii)
Multiplying (i) by 3 and (ii) by 2 and subtracting, we get
3(2X + 3Y) – 2(3X + 2Y) = 3 =
2
4
3
−2
G – 2=
0
1
⇒
6 9
−4
5Y = =
G–=
12 0
2
⇒
10
5Y = =
10
⇒
Y=
2
G
−5
4
6+4
G==
10
12
−
−2
5
G
10
1 10
=
5 10
5
2
G==
10
2
1
G
2
Substituting Y in (i) we get
2X = =
2
4
2
3
G – 3=
2
0
==
2
4
3
6
G–=
0
6
==
−4
−2
1 −4
=
2 −2
⇒
X=
Hence,
X==
−2
−1
1
G
2
3
2−6
G==
6
4−6
3−3
G
0−6
0
G
−6
0
−2
G==
−6
−1
0
2
G; Y = =
2
−3
0
G
−3
1
G
2
18. Let A be first term and R be common ratio of a GP.
Then a = ARp–1, b = ARq–1, c = ARr–1
Consider
log a
Δ = log b
log c
p 1
q 1
r 1
log AR p − 1
= log AR q − 1
log AR r − 1
8
p 1
q 1
r 1
9−4
G
0 + 10
log A + (p − 1) log R
= log A + (q − 1) log R
log A + (r − 1) log R
p 1
q 1
r 1
log A
= log A
log A
p 1 (p − 1) log R
q 1 + (q − 1) log R
r 1 (r − 1) log R
p 1
q 1
r 1
1
= log A 1
1
p 1
p−1
q 1 + log R q − 1
r 1
r−1
p 1
q 1
r 1
p
= log A × 0 + log R q
r
= 0 + log R × 0 = 0
19. Consider
=
=
=
=
=
=
y
1
a2 –b
x2
(x2 + a2) (x2 + b2)
a2 − b
1
a2 − b
1
a2 − b
(x2 + a2) (x2 + b2)
=y
2
>y
2
x2
x2 + b2
a2 − b2
1
a2 − b2
y
(x2 + b2) − b2
(x2 + b2)
= y e1 −
2
1
dx −
=x −
b2
x2 + b
[Performing C1 → C1 + C3]
dx
x2 [(x2 + a2) − (x2 + b2)
y
2
1
p 1
q 1
r 1
dx
x2
x2 + a2
dx −
y
dx G
(x2 + a2) − a2
(x2 + a2)
o dx − y e1 −
2
a2
x2 + a2
dx H
o dx G
a2
x
b2
x
tan − 1 − x +
tan − 1 G + c
a
a
b
b
= a tan − 1
x
x
− b tan − 1 G + c
a
b
OR
Consider
y
1
4
x +1
dx
=
1
2
1 (1 + x2) + (1 – x2)
dx = y
dx
y
2 x4 + 1
2
x4 + 1
=
1 x2 + 1
dx –
=y
2 x4 + 1
y
x2 – 1
x4 + 1
dx G
...(i)
9
Consider
y
x2 + 1
4
x +1
1+
dx =
=
y
Consider
y
4
x +1
x +
y
=
x dx =
1
t +2
1
2
2
y
t
=
2
1
2
x dx =
1
x2
1
2
y
1
x2
1 2
cx – m + 2
x
dx
dt
tan –1
x +
1+
x2
2
1–
dx =
2
1
y
=
x 2 –1
2
1
1
2
tan –1
1–
y
1
=t
x
1
⇒ e1 + o dx = dt
x2
Let x –
1
2
x = 1 tan − 1 x − 1
2
2
2x
x–
1
x2
1 2
cx + m – 2
x
dx
dt
1
=t
x
1
⇒ e1 – o dx = dt
x2
Let x +
t –2
1
t– 2
=
log
2 2
t+ 2
1
x+ – 2
1
x2 + 1 − 2 x
1
x
=
log
=
log
1
2 2
2 2
x2 + 1 + 2 x
x+ + 2
x
Substituting from (ii) and (iii) in (i), we get
y
1 1
x2 – 1
1
x2 + 1 – 2 x
dx = =
tan –1
–
log
G+ c
2 2
2x 2 2
x4 + 1
x2 + 1 + 2 x
1
20. Given relation is R = {(P1, P2) ∈ A × A : P1 and P2 have same number of sides}
For reflexive: For P ∈ A
(P, P) ∈ R ⇒ P, P have same number of sides, true.
Hence, reflexive.
For symmetric: For P, Q ∈ A
(P, Q) ∈ R ⇒ P and Q have same number of sides
⇒ Q and P have same number of sides
⇒ (Q, P) ∈ R
As (P, Q) ∈ R ⇒ (Q, P) ∈ R for P, Q ∈ A
Hence, symmetric.
For transitive: For P, Q, T ∈ A
Let (P, Q) ∈ R and (Q, T) ∈ R
10
...(ii)
...(iii)
⇒
P and Q have same number of sides and Q and T have same number
of sides
⇒
P and T have same number of sides
⇒
(P, T) ∈ R
As (P, Q) ∈ R, (Q, T) ∈ R ⇒ (P, T) ∈ R for P, Q, T ∈ A .
Hence, transitive.
As relation R is reflexive, symmetric and transitive.
Hence, relation R is an equivalence relation.
Set of all elements in set A related to right triangle T with sides 3, 4 and 5.
⇒ These elements should have same number of sides as right angled triangle has, i.e. 3
∴ Set of all element in set A related to right triangle T is set of all triangles.
OR
Given set {0, 1, 2, 3, 4, 5} and binary operation
a * b = (a + b) mod 6, where (a + b) mod 6 is remainder obtained on dividing (a + b) by 6.
Operation table for *
*
0
1
2
3
4
5
0
0
1
2
3
4
5
1
1
2
3
4
5
0
2
2
3
4
5
0
1
3
3
4
5
0
1
2
4
4
5
0
1
2
3
5
5
0
1
2
3
4
From table we notice that a * 0 = a and 0 * a = a
Hence, 0 is identity for this operation.
For inverse, we notice.
5 * 1 = 0 ⇒ 5 * (6 – 5) = 0
4 * 2 = 0 ⇒ 4 * (6 – 4) = 0
3 * 3 = 0 ⇒ 3 * (6 – 3) = 0
2 * 4 = 0 ⇒ 2 * (6 – 2) = 0
1 * 5 = 0 ⇒ 1 * (6 – 1) = 0
In general
a * (6 – a) = 0
⇒ 6 – a is inverse of a.
11
21. Let square of side x cm be cut off from each corner to make a box.
Then
l = (45 – 2x) cm, b = (24 – 2x) cm, h = x cm ...(i)
x
x
x
x
24 cm
Volume of the box (V) = x(45 – 2x) (24 – 2x)
V = 2x(45 – 2x)(12 – x)
x
x
x
Differentiating both sides w.r.t. x, we get
x
45 cm
dV
= 2[x(45 – 2x)(–1) + x(12 – x) (–2) + (45 – 2x) (12 – x) (1)]
dx
= 2[–45x + 2x2 – 24x + 2x2 + 540 – 69x + 2x2]
= 2[6x2 – 138x + 540] = 12[x2 – 23x + 90]
= 12(x – 18) (x – 5)
For maximum V,
⇒
dV
= 0 ⇒ x – 18 = 0 or x – 5 = 0
dx
x = 18 (rejected) [from (i)] or x = 5
d2 V
dx2
d2 V
dx2
G
= 12[(x – 18) ⋅ 1 + (x – 5) ⋅ 1] = 12 [2x – 23]
= 12[10 – 23] < 0
x=5
Hence, for x = 5, volume is maximum.
Hence, a square of side 5 cm must be cut off from each corner to make a box of maximum
volume.
22. Given reigon {(x, y) : x2 ≤ y ≤ |x|}
Corresponding inequations are y ≥ x2, y ≤ |x|
Plotting the graph of inequations we notice we have to find the shaded area.
we notice both the curves y = x2 and y = |x|
are symmetrical with respect to y-axis as both the
functions are even with respect to x.
Y
∴ Area = 2 × area in first quadrant.
For first quadrant, equations are y = x , y = x,
(x > 0)
y=
2
y=
–x
x2
y=
Eliminating y, we get x2 = x ⇒ x(x – 1) = 0
⇒ x = 0, 1
∴
Area = 2 y (x − x2) dx = 2 =
1
0
1
x 2 x3
− G
2 3 0
1 1
1 1
= 2 ;c − m − 0E = 2 × = sq units.
2 3
6 3
12
–1
0
1
X
x
x2
23. Let ellipse be
y2
= 1, (a > b)
a2 b2
Differentiating both sides w.r.t. x, we get
1
a2
⇒
+
1
(2x) +
b
c 2y $
2
...(i)
Y
F2
dy
m=0
dx
F1
0
X
y dy
x
y dy
b2
⇒ $
=–
=–
x dx
b2 dx
a2
a2
Again differentiating both sides w.r.t. x, we get
x) y $
d2 y
dx2
+c
dy 2
dy
m 3 – cy $
m1
dx
dx
x2
⇒ xy
d2 y
dx
2
+ xc
=0
dy 2
dy
= 0 is the required differential equation.
m –y
dx
dx
OR
Consider equation (1 + y + x2y)dx + (x + x3)dy = 0
⇒ {1 + y(1 + x2)}dx + x(1 + x2)dy = 0
⇒ x(1 + x2)
⇒
dy
dx
=
dy
dx
= –{1 + y(1 + x2)}
−1
2
−
y
x
x (1 + x )
y
−1
⇒
+ =
dx x
x (1 + x2)
dy
Here, P(x) =
1
−1
, Q(x) =
x
x (1 + x2)
Integrating factor = e y
Solution is, (I.F)y =
x.y =
1
dx
x
= elog x = x
y {(I $ F) Q (x)} dx
−1
y x$
x (1 + x2)
1
= −y
dx
1 + x2
xy = – tan–1x + c
dx
...(i)
When x = 1, y = 0
⇒ 0 = – tan–11 + c ⇒ c =
π
4
13
Substituting in (i), we get
xy = – tan–1x +
π
is the required solution.
4
x−1 3−y z+1
=
=
5
2
4
x−1 y−3 z+1
or
=
=
5
4
−2
Vector equation of the line is
24. Given line is
r = ^it + 3tj − kth + λ^5it − 2tj + 4kth
...(i)
As lines are parallel, so direction vector of required line is ^5it − 2tj + 4kth and line passes
through the point (3, 0, –4), i.e. point with position vector ^3it − 4kth
∴ Vector equation of the line is
r = ^3it − 4kth + μ^5it − 2tj + 4kth
...(ii)
From line (i), a1 = it + 3tj − kt , b = 5it − 2tj + 4kt
From line (ii), a2 = 3it − 4kt , b = 5it − 2tj + 4kt
Distance between parallel lines =
(a2 − a1) × b
...(iii)
b
a2 − a1 = 3it − 4kt − it − 3tj + kt = 2it − 3tj − 3kt
it
^ a2 − a1h × b = 2
5
^ a2 − a1h × b =
b =
kt
− 3 = it (− 18) − tj (23) + kt (11) = − 18it − 23tj + 11kt
4
tj
−3
−2
324 + 529 + 121 =
25 + 4 + 16 =
974
45
974
units
45
∴ Distance between parallel lines =
{from (iii)}
1 ball
25.
I
6R+5B
II
5R+8B
Case I: When ball transferred is red.
P` R II j =
5
13
Number of balls in bag I = (6 + 1) R + 5 B
P` B I j =
5
12
∴ Probability in this case =
14
5
5
×
13 12
...(i)
Case II: When ball transferred is blue.
P` B II j =
8
13
Number of balls in bag I = 6 R + ( 5 + 1) B = 6 R + 6 B
P` B I j =
6
12
∴ Probability in this case =
8
6
×
13 12
...(ii)
∴ Probability of drawing a blue ball from bag I when one ball is transferred from
5
5
8
6
[From (i) & (ii)]
bag II to bag I is
×
×
+
13 12 13 12
25 + 48
73
=
=
.
156
156
26.
Cake I (x)
Cake II (y)
Flour
Fat
200 g
25 g
100 g
50 g
≤ 5 kg
≤ 1 kg
Let x cakes of type I and y cakes of type II are made.
Then LPP is
Y
50
To maximise Z = x + y
40
Subject to constraints
30
x ≥ 0, y ≥ 0
20
200 x + 100 y ≤ 5000 ⇒ 2x + y ≤ 50
10
25 x + 50 y ≤ 1000 ⇒ x + 2y ≤ 40
0
20
A(25, 0)
30 40
x 50
+
2y
y
=
X
=
40
50
Z=x+y
10
+
Point
B(20, 10)
2x
Plotting the graph of inequations we notice shaded portion
is feasible solution. Possible points for maximum Z are
A(25, 0), B(20, 10), C(0, 20)
C(0, 20)
Value
A(25, 0)
25 + 0
25
B(20, 10)
20 + 10
30
C(0, 20)
0 + 20
20
← Maximum
We notice for B(20, 10), i.e. x = 20, y = 10, Z is maximum.
Hence, 20 cakes of type I and 10 cakes of type II must be made for maximum number of
cakes within given constraints.
15