Homework #1 Solutions 3.1 a. Let X1 =the number of hours of process 1 used X2 =the number of hours of process 2 used The objective is to minimize the cost of producing the three products A, B and C. The constraints are on the demand of each product. Min 4X1 + X2 subject to 3X1 + X2 ≥ 10 (A’s demand) X1 + X2 ≥ 5 (B’s demand) X1 ≥ 3 (C’s demand) X1 , X2 ≥ 0 (nonnegativity) b. From the graph we can see that the isocost line just leaves the feasible region where the demand for C and B intersect. The point of intersection is (3,2). The cost at that point is 14. That is the minimum cost. Graphical Solution to 3.1

10 9 8 7 6 5 4 3 2 1 0 0 1 2 3 X1 4 5 6

A B C Isocost=14

3.7

a. Let X1 =the number of type 1 trucks produced X2 =the number of type 2 trucks produced The objective is to maximize the profit of producing the two truck types. The constraints are capacities of the paint and assembly shops. Max 300X1 + 500X2 subject to 7X1 + 8X2 ≤ 5600 (Paint Shop) 4X1 + 5X2 ≥ 6000 (Assembly Shop)

X2

X1 , X2 ≥ 0, integers (nonnegativity & integrality) 3.10 We want to minimize production and inventory costs while meeting demand. Only half of a week’s demand can be filled from that week’s production. I assume that the other half is available to meet the following weeks’ demand. Let pi=the number produced dur ing week i, i=1, 2, 3 Ai= the number available to meet demand during week i, i=1, 2, 3 Ii=the number in inventory at the end of week i, i=1, 2, 3

Min 13p1 + 14p2 + 15p3 + 2(I1 + I2 + I3 ) subject to A1 = 5 + .5p1 (Number available to meet demand for week 1) A1 ≥ 20 (We must meet demand for week 1) I1 = A1 -20 + .5 p1 (Balance constraint for week 1) A2 = I1 + .5p2 (Number available to meet demand for week 2) A2 ≥ 10 (We must meet demand for week 2) I2 = A2 -10 + .5 p2 (Balance constraint for week 2) A3 = I2 + .5p3 (Number available to meet demand for week 3) A2 ≥ 15 (We must meet demand for week 2) I3 = A3 -15 + .5 p3 (Balance constraint for week 2) All...

...Practical 7
Title:
Introduction to preparing standardized solutions and making dilutions
Introduction:
A solution is a mixture of two or more substances uniformly dispersed through a single phase, so that the mixture has the same composition throughout. A solution may be a solid dissolved in a liquid. A solution is comprised of a solvent and a solute, where: the solvent is the liquid in which the solid dissolves (Water is the most...

...permanganate (KMnO4) solution by finding its absorbance through the use of spectrophotometer. The preparation of four known concentration of KMnO4 was done namely, 2.00×10-4M, 1.50×10-4M, 1.00×10-4M, 5.00×10-5M, respectively and is to be place on the spectrophotometer with the unknown and distilled water for the determination of each concentration’s absorbance. As the concentration is proportional with the absorbance of the solution, to determine the concentration...

...Depression in Solutions
The freezing point of pure water is 0°C, but that melting point can be depressed by the adding of a solvent such as a salt. The use of ordinary salt (sodium chloride, NaCl) on icy roads in the winter helps to melt the ice from the roads by lowering the melting point of the ice. A solution typically has a measurably lower melting point than the pure solvent.
The following figures were found in a published report, but have not been checked...

...
Purpose
The purpose of this experiment is to find out the molarity of each solution while finding out how many grams of sodium chloride we have.
Materials
1. Wire gauze
2. tongs
3. Burner
4. balance
5. Matches
6. hot hands
7. ring stand
8. ring clamp
9. graduated cylinder
10. evaporating dish
Procedure
Steps:
1. Mass the evaporating dish
2. Record how many mL of solution you add into the evaporating dish then mass them...

...Directions Using the formula for dilutions, write the recipe to make 100 mL of a new solution of each solute based on having as much of the initial solution as you need. V1 M1 V2M2 SoluteFormula of SoluteInitial Concentration (M)Final Concentration (M)Recipe For 100 mL of New SolutionSodium hydroxideNaOH1.0 M0.27 M(1.0M)(VNaOH) (0.27 M)(100 mL) VNaOH 27 mL Dilute 27 mL of 1.0 M NaOH to 100 mL.Potassium...

...Synergetic Solutions Report
This paper will discuss the internal and external forces of change for Synergetic Solutions. The report will show the factors that the leader in this organization needs to consider to implement a change strategy successfully and it will describe at least two change models the leader might employ. And we will evaluate the communication necessary to implement change using these models. The paper will show what kinds of resistance might...

...Abstract
This laboratory involved utilizing equipment to dilute a sugar water solution. It also created solutions containing varying levels of concentrations and densities. Equations were used to figure the molecular weight of the sugar, and the number of moles of sugar in the volumetric flask. There was also an equation to figure the Molarity, as well. As a result of the experiment, I now have a better understanding of the density of a concentration, and what...

...Mixtures and Solutions can often become confusing because solutions are mixtures, but not all mixtures are solutions. A mixture can either be homogeneous or heterogeneous. A homogeneous mixture is where the mixture’s components are distributed uniformly within the mixture. A heterogeneous mixture is where the components are not uniform. Mixtures can either be miscible or immiscible, the difference being whether or not the mixture forms a homogeneous...