Question

4:07 MM. ONS 84% Mathematica... E Saved on device D. I pore 4.970 interest compounded quarterly. c. If it bore 5% interest compounded continuously 10) How much of a 10% salt solution must be mixed with a 20% salt solution to obtain 100 liters of a 15% salt solution? (4.1, pp. 201-203) 11) A factory turns out tables and chairs. We have the following information as shown on the table: (4.1,5.3) bor-Hours per Unitu m Labor Hour ment Available per di able Chair 82 bly 400 120 2 1 per unit $90 $25 a. If x is the number of tables and y the number of chairs produced, write an inequality of the amount of hours can be used in the Assembly and the Finishing departments b. Graph the inequalities in (a). c. Find the coordinates of the corners of the graph drawn in (b). d. Use the coordinates of the corners found in (c) to calculate the maximum profits under the constraints give.

Answer 1

o la X= litres of 10% solution ie. O.121 Y litses of 20'/. Solecion iie. 0.244, i. To make 15% of 100 liter ie .0.154100) .: 0.12, + 0.24, =015 U00) 0.12, +0.2X2-15 - 0 The total 100 litres, so.. Xx+x2=100 - © soling, both equations, we get, 0.12, +02x2=15 -0.12,-0.11,2-10 0.122 - 5 |x₂²50 But x2 = 50 inq@, weget, X: +507100 .: B=50 : 50 litses of 10% and 20% solution each a) let x = No. of tables y = No.of chairs Contraints: 8x+ 2y <400 (Assembly longtouint i 400 available) 2x+y =120 C Finishing constaunt!+ 20 Quailable) Objective function: Mak 2 =90x+ 25y

20 30 40 50

. c) Cosnes Points of the feasible region from graph, 0 C0,0) , A (50,0), BCHO, 40), C (0,120)

d) In this, I used the x2 variable instead of y variable.

Objective function value Z= 9021 + 25x2 90(0) + 25(0) = 0 The value of the objective function at each of these extreme points is as follows: Extreme Point Coordinates Lines through Extreme Point (21,22) 0(0,0) 3 + x1 > 0 4 + x2 > 0 1 + 8x1 + 2x2 < 400 A(50,0) 4 → 22 > 0 B(40,40) 1 +821 + 2x2 < 400 2 2x1 + x2 < 120 C(0,120) 2 + 2x1 + x2 < 120 3 → 21 20 90(50) + 25(0) = 4500 90(40) + 25(40) = 4600 90(0) + 25(120) = 3000 The maximum value of the objective function Z = 4600 occurs at the extreme point (40, 40). Hence, the optimal solution to the given LP problem is : x1 = 40, x2 = 40 and max 2 = 4600.