Mid-Chapter Quiz: Lessons 3-1 through 3-4 Solve each system of equations. 4.
SOLUTION: Multiply the first equation by 2 and add with the second equation.
Substitute 10 for x in the first equation and solve for y.
The solution is (10, 6). 15. CARPENTRY Cal’s Carpentry makes tables and chairs. The process involves some carpentry time and some finishing time. The carpentry times and finishing times are listed in the table below.
Cal’s Carpentry can work for a maximum of 108 carpentry hours and 20 finishing hours per day. The profit is $35 for a table and $25 for a chair. How many tables and chairs should be made each day to maximize profit? a. Using c for the number of chairs and t for the number of tables, write a system of inequalities to represent this situation. b. Draw the graph showing the feasible region. c. Determine the number of tables and chairs that need to be made to maximize profit. What is the maximum profit? SOLUTION: a. Let c and t be the number of chair and table. Optimization function:
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Mid-Chapter Quiz: 3-1 through 3-4 The solution is (10,Lessons 6). 15. CARPENTRY Cal’s Carpentry makes tables and chairs. The process involves some carpentry time and some finishing time. The carpentry times and finishing times are listed in the table below.
Cal’s Carpentry can work for a maximum of 108 carpentry hours and 20 finishing hours per day. The profit is $35 for a table and $25 for a chair. How many tables and chairs should be made each day to maximize profit? a. Using c for the number of chairs and t for the number of tables, write a system of inequalities to represent this situation. b. Draw the graph showing the feasible region. c. Determine the number of tables and chairs that need to be made to maximize profit. What is the maximum profit? SOLUTION: a. Let c and t be the number of chair and table. Optimization function:
Constraints:
b.
c. The vertices of the feasible region are (0, 0), (0, 20), (34, 3) and (36, 0). Substitute the points (0, 0), (0, 20), (34, 3) and (36, 0) in the function f (x, y) = 25c + 35t.
The maximum value is 955 at (34, 3). Therefore, 34 chairs and 3 tables will give a maximum profit of $955. Graph each system of inequalities. Name the coordinates of the vertices of the feasible region. Find the Page 2 maximum and the minimum values of the given function.
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The maximum value is 955 3-1 at (34, 3). 3-4 Mid-Chapter Quiz: Lessons through Therefore, 34 chairs and 3 tables will give a maximum profit of $955. Graph each system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum and the minimum values of the given function.
17.
SOLUTION:
The vertices of the feasible region are (–4, –3), (0,5) and (2, –3). Substitute the points (–4, –3), (0,5) and (2, –3) in the function f (x, y) = 4x – 3y.
The maximum value is 17 at (2, –3) and the minimum value is –15 at (0, 5). 19. GEOMETRY An isosceles trapezoid has shorter base of measure a, longer base of measure c, and congruent legs of measure b. The perimeter of the trapezoid is 58 inches. The average of the bases is 19 inches and the longer base is twice the leg plus 7. a. Find the lengths of the sides of the trapezoid. b. Find the area of the trapezoid. SOLUTION: a. The equation representing the situation is
Simplify the second equation. Substitute 38 for a + c in the first equation and solve for b.
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Substitute 10 for b in the third equation and solve for c.
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Simplify the second equation. Mid-Chapter Quiz: Lessons 3-1 through 3-4 Substitute 38 for a + c in the first equation and solve for b.
Substitute 10 for b in the third equation and solve for c.
Substitute 27 for c in the second equation and solve for a.
Therefore, the measures of the sides a, b and c are 11 in., 10 in. and 27 in. respectively.
b. The area of a trapezoid is
.
The bases and the height of the trapezoid is 11 in., 27 in., and 6 in. Substitute the values in the area formula and simplify.
2
Therefore, area of the trapezoid is 114 in .
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