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MAT1050 CORE ALGEBRA EXAM Q & A WITH RATIONALES 2024. $12.49   Add to cart

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MAT1050 CORE ALGEBRA EXAM Q & A WITH RATIONALES 2024.

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MAT1050 CORE ALGEBRA EXAM Q & A WITH RATIONALES 2024.

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  • January 26, 2024
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  • 2023/2024
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MAT1050




Core Algebra

LATEST EXAM w/ RATIONALES



2024

,PART A:

1. A company produces two types of widgets, A and B.
The production cost of widget A is $5 per unit and the
selling price is $8 per unit. The production cost of
widget B is $7 per unit and the selling price is $10 per
unit. The company can produce at most 1000 widgets
per day, and the demand for each type of widget is at
least 300 units per day. How many units of each type of
widget should the company produce to maximize its
profit?
a) 300 units of A and 700 units of B
b) 400 units of A and 600 units of B
c) 500 units of A and 500 units of B
d) 600 units of A and 400 units of B
Answer: b) 400 units of A and 600 units of B
Rationale: The profit function is P = 3A + 3B, where A
and B are the number of units of widget A and B,
respectively. The constraints are A + B ≤ 1000, A ≥
300, and B ≥ 300. Using the graphical method, we can
find the feasible region and the corner points. The
corner points are (300, 700), (400, 600), (700, 300), and
(1000, 0). Evaluating the profit function at each corner
point, we find that P is maximized when A = 400 and B
= 600, with a maximum profit of $3600.

2. A farmer has a rectangular field with a perimeter of
200 meters. He wants to divide the field into two plots
by building a fence parallel to one of the sides. What are

, the dimensions of the field that will maximize the area
of one of the plots?
a) 25 m by 75 m
b) 40 m by 60 m
c) 50 m by 50 m
d) None of the above
Answer: c) 50 m by 50 m
Rationale: Let x be the length of the side parallel to the
fence, and y be the width of the field. Then the area of
one plot is A = xy/2. The perimeter constraint is 2x + 2y
= 200, or y = 100 - x. Substituting y into A, we get A =
x(100 - x)/2. To maximize A, we take the derivative and
set it equal to zero: dA/dx = (100 - x)/2 - x/2 = (100 -
2x)/2 = 0. Solving for x, we get x = 50. Then y = 100 - x
= 50. So the dimensions that maximize the area are 50
m by 50 m.

3. A manufacturer produces two models of bicycles,
standard and deluxe. The standard model requires 4
hours of assembly time and 2 hours of painting time.
The deluxe model requires 6 hours of assembly time
and 3 hours of painting time. The manufacturer has a
total of 240 hours of assembly time and 120 hours of
painting time available per week. The profit on each
standard model is $80 and on each deluxe model is
$120. How many bicycles of each model should the
manufacturer produce per week to maximize its profit?
a) 20 standard and 20 deluxe
b) 24 standard and 16 deluxe

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