Think Break #8 Repeat the previous problem, but now use a milk price of Milk $10/cwt M = – 25.9 + 2.56G + 1.05H – 0.00505G2 – 0.00109H2 – 0.00352GH Prices:

Slides:



Advertisements
Similar presentations
With Rational Functions
Advertisements

Solving Equations = 4x – 5(6x – 10) -132 = 4x – 30x = -26x = -26x 7 = x.
Solving Systems of three equations with three variables Using substitution or elimination.
7.8 Equations Involving Radicals. Solving Equations Involving Radicals :  1. the term with a variable in the radicand on one side of the sign.  2. Raise.
Notes Over is what percent of 140?
3.5 Solving systems of equations in 3 variables
Write and graph a direct variation equation
Systems Warm-Up Solve each linear system. 1.x + 7 = y -4x + 2 = y 2.x + 2y = 10 3y = 30 – 2x.
Substitute 0 for y. Write original equation. To find the x- intercept, substitute 0 for y and solve for x. SOLUTION Find the x- intercept and the y- intercept.
Substitute 0 for y. Write original equation. To find the x- intercept, substitute 0 for y and solve for x. SOLUTION Find the x- intercept and the y- intercept.
Solving System of Linear Equations
Multiple Input Production Economics for Farm Management AAE 320 Paul D. Mitchell.
Substitution Method: 1. Solve the following system of equations by substitution. Step 1 is already completed. Step 2:Substitute x+3 into 2 nd equation.
Solving Linear Systems by Substitution O Chapter 7 Section 2.
Solving a system of equations by adding or subtracting.
3.6 Solving Absolute Value Equations and Inequalities
1.2 The Method of Substitution. Intersection of Lines Yesterday we talked about solving when two lines intersect… – How did we do that? – Are they equal.
Solving by Elimination Example 1: STEP 2: Look for opposite terms. STEP 1: Write both equations in Standard Form to line up like variables. STEP 5: Solve.
Chapter 3 –Systems of Linear Equations
Section 5.3 Solving Systems of Equations Using the Elimination Method There are two methods to solve systems of equations: The Substitution Method The.
3-2 Solving Systems Algebraically: Substitution Method Objective: I can solve a system of equations using the substitution method.
Linear Programming The Table Method. Objectives and goals Solve linear programming problems using the Table Method.
Elimination Method Day 2 Today’s Objective: I can solve a system using elimination.
Warm Up 04/01 Practice CRCT Problem: Solve the following system of equations using the substitution method: 2x − 3y = − 1 y = x − 1.
The Substitution Method Objectives: To solve a system of equations by substituting for a variable.
SOLVING SYSTEMS OF EQUATIONS BY SUBSTITUTION. #1. SOLVE one equation for the easiest variable a. Isolated variable b. Leading Coefficient of One #2. SUBSTITUTE.
Notes 6.5, Date__________ (Substitution). To solve using Substitution: 1.Solve one equation for one variable (choose the variable with a coefficient of.
Solve a two-step equation by combining like terms EXAMPLE 2 Solve 7x – 4x = 21 7x – 4x = 21 Write original equation. 3x = 21 Combine like terms. Divide.
Solving equations with variable on both sides Part 1.
Algebra Review. Systems of Equations Review: Substitution Linear Combination 2 Methods to Solve:
SOLVING SYSTEMS OF EQUATIONS BY SUBSTITUTION PRACTICE PROBLEMS.
Solving Systems by Substitution (isolated) Solving Systems by Substitution (not isolated)
Textbook pages 69 & 73 IAN pages 91 & 95. Each side of the equation must be balanced. We balance an equation by doing the same thing to both sides of.
Warmups – solve using substitution
How to Write an Equation of a Line Given TWO points
7.3 Solving Equations Using Quadratic Techniques
Elimination Method Day 1
Solve Systems of Equations by Elimination
Multiple Input Production Economics for Farm Management
Solving Systems using Substitution
Pre-Algebra Unit 5 Review
3-1 HW:Pg #4-28eoe, 30-48e, 55, 61,
Think Break #3 N lbs/A Yield bu/A MP VMP
Solve an equation by multiplying by a reciprocal
3-2 Solving Systems Algebraically: Substitution Method
Think Break #10 You work for UWEX and have data on several farms in your seven county district You look at all farms with similar sized milking parlors.
Think Break #11 These are the Think Break #10 data (FC = $10,000)
Section 11.2: Solving Linear Systems by Substitution
Warm-Up Solve the system by substitution..
Algebra: Graphs, Functions, and Linear Systems
Think Break #7 Grain (lbs) Hay MRTS price ratio
WARMUP 1. y > – x – 4 y < 2x + 2.
Section 2 – Solving Systems of Equations in Three Variables
Solve a system of linear equation in two variables
السيولة والربحية أدوات الرقابة المالية الوظيفة المالية
CHAPTER SEVEN System of Equations.
3.5 Solving systems of equations in 3 variables
Solving Two-Step Equations
Think Break #9 (Review) Steers Beef MP VMP
Solving Systems of Equation by Substitution
Solving One and Two Step Equations
Solving One Step Equations
Bell work Week 20.
Systems of Equations Solve by Graphing.
Warm-Up Solve the system by graphing..
RELATIONS & FUNCTIONS CHAPTER 4.
75 previous answer What is of 37.5? ? go to.
Notes: 2-1 and 2-2 Solving a system of 2 equations:
75 previous answer What is of 60? ? go to.
N-player Cournot Econ 414.
Presentation transcript:

Think Break #8 Repeat the previous problem, but now use a milk price of Milk $10/cwt M = – 25.9 + 2.56G + 1.05H – 0.00505G2 – 0.00109H2 – 0.00352GH Prices: Grain $3/cwt, Hay $1.50/cwt What are the profit maximizing inputs?

Think Break #8: Answer p = 0.1(– 25.9 + 2.56G + 1.05H – 0.00505G2 – 0.00109H2 – 0.00352GH) – 0.03G – 0.015H FOC’s pG = 0.1(2.56 – 0.0101G – 0.00352H) – 0.03 = 0 pH = 0.1(1.05 – 0.00218H – 0.00352G) – 0.015 = 0 2.56 – 0.0101G – 0.00352H = 0.03/0.10 = 0.3 1.05 – 0.00218H – 0.00352G = 0.015/0.10 = 0.15

Think Break #8: Answer Solve FOC1 for H 2.56 – 0.0101G – 0.00352H = 0.3 0.00352H = 2.26 – 0.0101G H = (2.26 – 0.0101G)/0.00352 H = 642 – 2.87G Substitute into FOC2 and solve for G 1.05 – 0.00218H – 0.00352G = 0.15 0.9 = 0.00218(642 – 2.87G) + 0.00352G 0.9 = 1.4 – 0.00626G + 0.00352G – 0.5 = – 0.00274G G = 0.5/0.00274 = 182.5 lbs/week

Think Break #8: Answer Substitute into equation for H H = 642 – 2.87G H = 642 – 523.8 = 118.2 lbs/week M = – 25.9 + 2.56G + 1.05H – 0.00505G2 – 0.00109H2 – 0.00352GH M = – 25.9 + 2.56(182.5) + 1.05(118.2) – 0.00505(182.5)2 – 0.00109(118.2)2 – 0.00352(182.5)(118.2) = 306 lbs/week p = 0.1(306) – 0.03(182.5) – 0.015(118.2) = $23.35/week