Mathematics Assignment

Mathematics Assignment
Student Name:
Instructor Name:
Course Number:
14th July 2019

Q1 a)
f (−12 )=f ( 12 )
2 x a3 ( 12 )3 +2 x 12 x a1=0
144 a3+ a1=0 ... ... .... (i)
f ( x)=a4 x4 + a3 x3 +a2 x2+ a1 x +a0
f '( x)=4 a4 x3 +3 a3 x2+ 2a2 x +a1
f ' ( 6 )=f ' (−6 ) =0 [ 6−6 are point of maxima ]
864 a4 +108 a3 +12 a2 + a1=0 ... ... ... .(ii)
−864 a4 +108 a3−12 a2 +a1=0 ... ... ... .(iii)
dF
dx =f (x )
24= ∫
−12
12
( a4 x4 +a3 x3 +a2 x2+ a1 x +a0 ) dx
208 a4 +720 a2 +15 a0=15 ... ... ...(iv)
From equations (i), (ii), (iii) and (iv) we have;
[ 0 1 0 144 0
0 1 12 108 864
0
15
1
0
−12 108 −864
720 0 62208 ] [a0
a1
a2
a3
]=
[ 0
0
0
15 ]
b) 62208 a4 +720 a2 +15 a0=15
a2=72 a4
15 a0 =15−62208 a4−720 a2
15 a0 =15−62208 a4−720∗72 a4

a0= 1
15 (15−114048 a4 )
c) f ( x )=a4 x4 +a4 72 x2 + 1
15 (15−114048 a4 )
a4 is free
d) 2 ∫
−10
10
( a4 x4 +a2 x2+ a0 ) dx <2
105
5 a4 + 103
3 a2 +10 a0 <1
a0=(1−7603.2 a4 )
Hence this inequality does not hold up.
2. a)
P(x, y) =Pbed(x) + Pbath(y) + P0
4Pbed + 3Pbath + P0=14...................................i
2Pbed + 2Pbath + P0=11.....................................ii
2Pbed + Pbath + P0=8........................................iii
ii-iii= Pbath =3
i-ii=2Pbed + Pbath=3.........................................iv
2Pbed + 3=3
Pbed=0
4Pbed + 3Pbath + P0=14
4(0) + 3(3) + P0=14
P0=14-9=5
P(x, y) = 3y + P0

E(x, y) =ebed(x) + ebath(y) + e0
4ebed +3 ebath + e0=15..................................m
2ebed +2 ebath + e0=10....................................n
2ebed +ebath + e0=9...........................................p
n-p= ebath=1
m-n=2ebed +ebath =5
2ebed +1 =5
ebed =2
4(2) +3(1) + e0=15
e0=15-8-3=4
E(x, y) =2x + y + 4
F(x, y) =fbed(x) + fbath(y) + f0
4fbed +3 fbath + f0=10..................................r
2fbed +2 fbath + f0=7....................................s
2fbed +fbath + f0=6...........................................t
s-t= fbath=1
r-s=2fbed +2 fbath=3
2fbed +1 =3
fbed =1
4(1) +3(1) + f0=10
f0=3
F(x, y) =x + y + 3
b)