MAT1050 Core Algebra LATEST EXAM w/ RATIONALES 2024

MAT1050

Core Algebra

LATEST EXAM w/ RATIONALES

2024

PART A:

• 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?

• 300 units of A and 700 units of B
• 400 units of A and 600 units of B
• 500 units of A and 500 units of B
• 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.

• 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?

• 25 m by 75 m
• 40 m by 60 m
• 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.

• 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?

• 20 standard and 20 deluxe
• 24 standard and 16 deluxe

• 30 standard and 10 deluxe

d) None of the above

Answer: b) 24 standard and 16 deluxe

Rationale: Let x be the number of standard models and

y be the number of deluxe models. Then the profit

function is P = 80x + 120y. The constraints are:

4x + 6y ≤ 240 (assembly time) 2x + 3y ≤

PART B:

• Which of the following is the correct equation for the

line passing through the points (2, 5) and (-3, 1)?

• y = -2x + 11
• y = -2x + 1
• y = 2x + 9
• y = 2x + 1

Answer: a) y = -2x + 11

Rationale: To find the equation of a line passing through

two points, we need to calculate the slope (m) first using the formula m = (y2 - y1) / (x2 - x1). Using the given points, we get m = (1 - 5) / (-3 - 2) = -4 / -5 = 4/5. Next, we substitute the slope and one of the points into the slope- intercept form y = mx + b to calculate the y-intercept (b).In this case, using the point (2, 5), we get 5 = (4/5)(2) + b, which simplifies to b = 11. Therefore, the equation of the line is y = -2x + 11.