MATH135 Mathematics IA Semester 2 2018 Assignment 3 Solution

This document presents a complete solution to MATH135 Assignment 3 from Macquarie University, Semester 2, 2018. The assignment covers several key areas, including linear programming and optimization, where the minimum value of a function is determined graphically and through the simplex method. The document also provides a solution to a pet food production optimization problem, maximizing profit under given constraints. Furthermore, the assignment incorporates calculus problems involving integration and the application of the Riemann sum. The solution includes detailed calculations, graphical representations, and MATLAB code for integration and Riemann sum calculations, demonstrating a comprehensive understanding of the mathematical concepts involved. The document concludes with an analysis of the upper Riemann sum, including its simplification and the calculation of its limit.

Macquarie University
Department of Mathematics
MATH135 Mathematics IA
Semester 2 2018
Assignment 3
Algebra
1. Minimizing
f ( x , y )=3 x+ 5 y
Subjected to the following constrains
3 x+ 2 y ≥ 36 ;3 x +5 y ≥ 45; x ≥ 0; y ≥ 0
Figure 1:
From figure 1 above:
The feasible points are ( 0,18 ) , ( 10,3 )∧ ( 15,0 )( 0 , 18 ) 3 ( 0 ) +5 ( 18 )=90 ( 10 , 3 ) 3 ( 10 )+5 ( 3 )=45
( 15 , 0 ) 3 (15 )+5 ( 0 )=45

Both points ( 10,3 )∧ (15,0 )have the same value, thus the minimum point can be either
( 10,3 ) ∨ ( 15,0 )
2. Let regular food to be x and premium food to be y. Here we need to maximize profit given
as:
Profit P=42 x +50 y
Subjected to the constraints:
2 x+2 y ≤8 3 x+ 5 y ≤ 15 x ≥ 0 ; y ≥0
Figure 2.
The points are ( 0 , 0 ) , ( 0 , 3 ) , ( 4,0 ) ∧(2.5,1.5) ( 0 , 0 ) 42 ( 0 ) +50 ( 0 ) =0 ( 0 , 3 ) 42 ( 0 ) +50 ( 3 )=150
( 2.5 , 1.5 ) 42 ( 2.5 ) + 50 ( 1.5 ) =180 ( 4 , 0 ) 42 ( 4 )+ 50 ( 0 )=168
Therefore, the pet food manufacturer should produce 2.5 regular food and 1.5 premium
food per day to get a profit of $ 180 per day
3. Maximize:
f ( x1 , x2 , x3 ) =14 x1+16 x2 +15 x3 Subjected to the constraints
x1+ x2 ,+ x3=10

−x1+ 2 x2−x3 ≤ 8
2 x1 +2 x2−4 x3 ≤14
x1 ≥ 0 ; x2 ≥ 0 ; x3 ≥ 0
Using Simplex Method
x1 x2 x3 s1 s2 s3 b
1 1 1 1 0 0 10
−1 2 −1 0 1 0 8
2 2 −4 0 0 1 14
−14 −16 −15 0 0 0 0

x1 x2 x3 s1 s2 s3 b
1 1 1 1 0 0 10
−1 2 −1 0 1 0 8
1 1 −2 0 0 1
2 7
2 0 −47 0 0 0 112
x1 x2 x3 s1 s2 s3 b
1 1 1 1 0 0 10
−1 2 −1 0 1 0 8
1 1 −2 0 0 1
2 7
2 0 −47 0 0 8 112

x1 x2 x3 s1 s2 s3 b
1 1 1 1 0 0 10
−1 2 −1 0 1 0 8
1 1 −2 0 0 1
2 7
0 0 0 0 6 4 582
Thus the solution that maximizes f ( x1 , x2 , x3 ) =14 x1+16 x2 +15 x3
x1=0 , x2=6 x3=4
Thus, the objective function f ( x1 , x2 , x3 ) =156
4. Formulate the optimization problem
Let the number of units to be produced in October, November and December to b x1 , x2 ,and
x3respectively. Then we need to minimize;
Minimize: f ( x1 , x2 , x3 ) =150 x1 +160 x2+160 x3
Subject to the constraints:
x1+ x2 +x3=280
x1 ≤ 100 , x2 ≤100 , x3 ≤ 100

x1 ≥ 70 , ( x1 −70 ) + x2 ≥ 90 , ( x2−90 ) +x3=120
Calculus
1. Given xsin ( πx ) =∫
0
x2
f ( t ) dt f ( t )= d
dt (∫
0
x2
f ( t ) dt )= d
dt ¿ ¿ sin ( πx ) +πxcos ( πx )
f ( 16 )=sin ( 16 π ) +16 π cos (16 π )¿ 0+16 π¿ 16 π
2. Given the figure f(x) shown. Areas are shown for three regions
Figure 3: graph of f(x)
a) Integration of given function shows the area under the curve, thus:
∫
0
3
f ( t ) dt=f ( x ) =f ( 3 ) =0∫
3
6
f ( t ) dt =∫
0
6
f ( t ) dt−∫
0
3
f ( t ) dt=f ( 6 ) −f (3)=0
∫
0
3
f ( t ) dt <∫
3
6
f ( t ) dt, is FALSE since ∫
0
3
f ( t ) dt =∫
3
6
f ( t )
b) Determining F(8)
F ( 8 )=∫
0
3
f ( x ) dx+∫
3
6
f ( x ) dx +∫
6
8
f ( x ) dx which is represented as the area under the
curve, Thus:
F ( 8 )=3+ (−5 )+4 ¿ 2
3. MATLAB

A..................................................................................................................................................1
B...................................................................................................................................................1
C...................................................................................................................................................1
D..................................................................................................................................................2
clc; clear;
format long
syms x
A
partA=cos(x);
partA_ans=int(partA)
partA_ans =
sin(x)
B
partB=x^5*cos(5*x);
partB_ans=int(partB)
partB_ans =
(24*cos(5*x))/3125 + (24*x*sin(5*x))/625 - (12*x^2*cos(5*x))/125 + (x^4*cos(5*x))/5 - (4*x^3*sin(5*x))/25 +
(x^5*sin(5*x))/5
C
partC=int(sqrt(x^5+x^3-1))
partC =

int((x^5 + x^3 - 1)^(1/2), x)
D
partD=x^3+exp(2*x);
partD_ans=vpaintegral(partD,x,[1 2])
partD_ans =
27.3545
Published with MATLAB® R2017a
Explanation to part C:
Writing int(sqrt(x^5 + x^3 - 1)) means
∫ √ x5 + x3 −1 dx, but since the integration MATLAB cannot perform, it returns the same
input, int((x^5 + x^3 - 1)^(1/2), x), the function to be integrated (x^5+x^3-1)^1/2 with
respect to x
4. Upper Riemann sum
a) Figure 4 below shows the upper Riemann sum of x2+1for n=5, n=10, n=20 and n=50

Figure 4: Upper Riemann sum for n=5, 10, 20 & 50
The interval [a b] is partitioned by pointsa< x1 < x2 <x3 <...< xn−1 <b, where the lengths
of the resulting intervals between the points are denoted ∆ x1 , ∆ x2 , ∆ x3 , … ∆ xn. If xi
¿
is an arbitrary point in thekth subinterval. Then:
Un =∑
i=0
n
f ( xk
¿ ) ∆ xi ∆ xi= ( a−b )
n = 2−0
n =2
n
f ( xk
¿
)=x2+1=( ( b−a ) i
n )2
+1= ( ( 2−0 ) i
n )2
+1= (2 i
n )2
+1∴ Un=∑
i=0
n
( ( 2 i
n )2
+1 ) ( 2
n )
b) Simplifying Un
Un =∑
i=0
n
(( 2 i
n )2
+ 1
) ( 2
n )¿ 2
n ∑
i=0
n
( 4 i2 +n2
n2 )= 2
n3 ∑
i=0
n
4 i2 +n2 ¿ 2
n3 ( n ( n+1 ) ( 2 n+1 )
6 ) + 10 n
3
¿ 2 ( n+1 ) ( 2 n+1 )
3 n2 + 10 n
3 lim
n → ∞
2 ( n+1 ) ( 2 n+1 ) +10 n2
3 n2 =
lim
n → ∞
2 (2 n2 +3 n+1 )
3 n2
¿ 8+20
6 = 28
6 = 14
3