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MTA611S - MATHEMATICS FOR AGRIBUSINESS - 1st Opp - JUNE 2022 |
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n Am I BI A u n IVER s I TY
OF SCIEnCE Ano TECHnOLOGY
FACULTYOF HEALTHAND APPLIEDSCIENCESAND NATURAL RESOURCES
DEPARTMENT OF AGRICULTURE & NATURAL RESOURCESSCIENCES
QUALIFICATION: BACHELOR OF SCIENCEIN AGRICULTURE
QUALIFICATION CODE: 07BASA
LEVEL: 6
COURSE CODE: MTA611S
COURSE NAME: Mathematics for Agribusiness
DATE: June 2022
PAPER: Theory
DURATION: 3 Hours
MARKS: 100
FIRSTOPPORTUNITY EXAMINATION QUESTION PAPER
EXAMINER(S)
MODERATOR:
Mr. Mwala Lubinda
Mr. Teofilus Shiimi
INSTRUCTIONS
1. Attempt all questions
2. Write clearly and neatly.
3. Number the answers clearly & correctly.
PERMISSIBLEMATERIALS
1. All written work MUST be done in blue or black ink
2. Calculators allowed
3. No books, notes and other additional aids are allowed
THIS QUESTION PAPER CONSISTS OF 7 PAGES (including this front page).
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QUESTION ONE
a. Give concise definitions of the following concepts related to functions:
i. Range
ii. Domain
b. Let f(a)
= (a 2 -
1
2a + 6) 2, compute f(l)
and f(-1).
c. Use interval notation to express the domain and range of the following function:
2k - 1
g(k) = kz - k
[MARKS]
(2)
(2)
(2)
(6)
d. Suppose you know that the production function that expresses the relationship
between table grapes output (q) and fertilizer application rate (x) is a quadratic
function that has: (i) maxima point and (ii) roots at O and 75. Based on the
provided information, answer the questions below
i. Derive the mathematical equation of the production function.
(3 )
ii. Find the critical point of the production function you have derived in d(i).
(5)
iii. Draw and label a graph that illustrates the production function. The graph
must clearly show the roots, maxima, and y-intercept points of the
(5)
production function.
TOTAL MARKS
[25]
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QUESTION TWO
a. Use mathematical expressions to concisely define the following concepts:
[MARKS]
i. Newton's Difference Quotient.
(2)
ii. Regular limit.
(2)
b. Briefly describe at least two algebraic approaches you would use to find the limit
(4)
of function at a given point, x = a.
c. Find:
i. I1. m-(-2-+h) 2 -4
h-+o
h
(2)
ii.
II.m
L-+1
~-1
L2
+2L-3
(3)
iii. h.m--,J-x-2-2
x-+ 6 X - 6
(5)
d. Find the equation of a straight-line that is tangent to the curve:
= y
q2 -
2q - 24
(7)
at q = 4.
TOTAL MARKS
[25]
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QUESTION THREE
a. Define the following concepts:
i. Partial derivative
ii. Cross derivative
b. Find the first derivative of the following function:
i. f (x) = (3x 4 - 5) 6
[MARKS]
(2)
(2)
(3)
- ii. f(L)
3~
(4)
c. Given a function:
z(x,y) = 3e7-2xy2
(6)
Find Zx, Zy and Zyz·
d. Optimize the following function by (i) finding the critical value(s) at which the
function is optimized and (ii) testing the second-order condition to distinguish
between a relative maximum or minimum.
(8)
q(x) = x 3 - 6x 2 - 135x + 4
TOTAL MARKS
[25]
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a. Find:
QUESTION FOUR
[MARKS]
i. I ~dt
(2)
ii. fo1c3x3- x + l) dx
(3)
iii. f 12x 2 (x 3 + 2)dx
(5)
b. In the manufacture of a product, fixed costs N$4000. If the marginal-cost
function is:
de
dq
=
250
+ 30q
-
9q 2
(5)
where c is the total cost (in dollars) of producing q kilograms of product. Find the
cost of producing 10 kilograms of the product.
c. To fill an order for 100 units of its product, a firm wishes to distribute production
between its two plants, plant 1 and plant 2. The total-cost function is given by:
c = f (q1,qz) = qf + 3q1 + 25q 2 + 1000
where q1 and q2 are the numbers of units produced at plants 1 and 2,
(10)
respectively. How should the output be distributed to minimize costs? {Hint:
assume that the critical point obtained corresponds to the minimum cost and the
constraint is q1 + q2 = 100).
TOTAL MARKS
[25]
THE END
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FORMULA
Basic Derivative Rules
C.n. 1:1~111 :\\.ulliph~ Rule !!_lc/(x)j- cf'(.,·)
d<
Po'.\\'t..1Ruh~. ~a(..,,·\\) - nx"·1
Suin Ruk ~lf(x)• g(x)I - j'(x)-• g'(x)
ax
Diffmncr Rnic ~[f(x)-g(x)I-
aY
f {x)- g'(x)
Prorluct Ruic 4a-x[/(x)~(x)]~
/(x)o'("x')- .v.(,x)/'(x)
Quc·1iei1<Rule --=t/-(fx) - g(.<)/'(.<)- f(x)g'(x)
a.xL g(.t) c
[!'(X)j'
-j; Ch,in Rulo· f(g(x)) - j te(x))e'(x)
Derivative Rules for Exponential Functions
-d ( e·• ) = e·•
dx
.!!_ (a' ) = a' In a
dx
.!!_ (e ''' >) = e' <'>g '( x )
dx
.!!_ ( a ' <'>) = 1n ( a ) a '<'> g '( x)
dx
Derivative Rules for Logarithmic Functions
-(dlnx)=-,x>O
dx
-ldn(g(x))
dx
I
x
= -u"-·'-( x)
g(x)
-(dlog
dx
0 x)=
--,xI
x In a
>0
d
-d
x
(log
0
g(x))=
0u '(x)
-
g(x)ln a
Basic Integration Rules
I. f a1fr=ax+C
• x•·•
2. x dr=-+C,
f n+ I
II~ -I
;dx C f 3.
= In 1-+'i
4. fe'd-r=e·'+C
5. fa'cfr=~+C
Ina
6. flnxdx=xlnx-x+C
Integration by Substitution
The following are the 5 steps for using the
integration by substitution metthod:
• Step 1: Choose a new variable u
• Step 2: Determine the value dx
• Step 3: Make the substitution
• Step 4: Integrate resulting integral
• Step 5: Return to the initial variable x
Integration by Parts
The formula for the method of integration by
parts is:
.ludv= 11.• v-_/vdu
There are three steps how to use this formula:
• Step 1: identify u and dv
• Step 2: compute u and du
• Step 3: Use the integration by parts
formula
Unconstrained
optimization:
Univariate
functions
The following are the steps for finding a solution
to an unconstrained optimization problem:
• Step 1: Find the critical value(s), such
that:
f '(a) = 0
• Step 2: Evaluate for relative maxima or
minima
o If f "(a) > 0 -1 minima
o If f "(a) > 0 -1 maxima
Unconstrained optimization: Multivariate
functions
The following are the steps for finding a
solution to an unconstrained optimization
problem:
Constrained Optimization
The following are the steps for finding a solution
to a constrained optimization problem using the
Langrage technique:
• Step 1: Set up the Langrage equation
Relarive 111axim11111Relarive 111ini11111111
I. f_,_.f=, ()
1.!,.,J;=0
2. f,_,.f,-v< Q
2. f,.,,f,.,. > 0
3. fu ·/:._>,.(f,_..f 3. f,.,·/,._>,.(f,,)2
• Step 2: Derive the First Order Equations
• Step 3: Solve the First Order Equations
• Step 4: Estimate the Langrage Multiplier
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Additionally:
• If fxx · [yy < Uxy) 2 , when fxx and [yy
have the same signs, the function is at
an inflection point; whenfxx and [yy
have different signs, the function is at
a saddle point.
• If fxx ·[yy = (fxy) 2 , the test is
inconclusive.