Download Solving Optimization Problems: Minimizing Costs of Rectangular Fences and more Study notes Calculus in PDF only on Docsity!
OPTIMIZATION
Optimization problems are word problems dealing with finding the maximum or minimum solutions to a problem. Examples of optimization problems are as follows:
- Given 20sq. ft. of cardboard, what are the dimensions of the biggest box that can be made?
- If you wanted to construct a cylindrical tin can that would hold 10 fluid ounces of a liquid, what are the dimensions that will use the least amount of material?
Before working optimization problems, a student must be aware of a couple of concepts from calculus.
- A student must know how to take the first derivative of a function of one variable.
- A student must be aware of the fact that in order to find the maximum or minimum points on a graph, one needs to take the first derivative of the function, set it equal to zero and solve for the unknown variable.
Let us look at an optimization problem. Be aware of the steps involved.
Example: A farmer wants to build a rectangular fence that will enclose 120 square feet for his dog Miff. The two long sides of the fence are to be made of Styrofoam at a cost of $5 per foot. The two shorter sides are to be made of wire at a cost of $6 per foot. What are the dimensions of the fence that will minimize cost?
First, let us draw a picture of the problem.
l
w w
l
Next, let us write out our Objective Equation. The object of the problem is to minimize cost, so our Objective Equation must be an equation that represents the total cost of the fence ( Cost Equation).
Cost = 5 l + 6 w + 5 l + 6 w
= 5 l + 5 l + 6 w + 6 w
= 10 l + 12 w
Our next step is to see if we are constrained in some way. If so, we must write an equation representing this constraint ( Constraint Equation ). We were told that we had to have an area of 120 square feet, so our constraint equation would be the equation for area.
A = wl = 120
Since the object of this problem is to minimize cost, we know that we will have to take the first derivative of the cost equation, set it equal to zero and solve for the unknown variable. Right now, however, we cannot take the first derivative of the cost equation because it is not in terms of one variable; therefore, our next step is to get the Objective Equation ( Cost Equation ) in terms of one variable. It is usually easiest to do this by using the Constraint Equation to solve for one of the variables in the objective equation
Since wl = 120, then w = 120/ l
and so substituting 120/ l for w in the Cost Equation , we get:
Cost = 10 l + 12(120/ l )
= 10 l + 1440/ l
= 10 l + 1440 l -
Now we are ready to take the first derivative of the cost equation
C = 10 l + 1440 l -
C' = 10 + (-1440) l -
C' = 10 - 1440 l -
Now we set the first derivative of the cost equation equal to zero and solve for l
0 = 10 - 1440 l –2^ Adding 1440 l -2^ to both sides we get
1440 l -2^ = 10
1440/ l^2 = 10 Multiplying by l^2 on both sides we get
1440 = 10 l^2
144 = l^2
Dividing by 10 on both sides we get
l = ±
l = 12
Since we cannot have a negative length, our only remaining answer is
- Determine Constraint P = 2 l + 2πr = 30 Equation
- Express the Objective A = πr^2 + 2 l r Equation in terms of (where l is 15 - πr as determined next) one variable. Since 2 l + 2πr = 30 2 l = 30 - 2πr l = 1/2 (30 - 2πr)
l =
Then A = πr^2 + 2(15 - πr)r = πr^2 + 30r - 2πr^2 A = 30r - πr^2
- Take the first A' = 30 - 2πr derivative of the 30 - 2πr = 0 Objective Equation , 30 = 2πr set it equal to zero, 30/2π = r and solve for your r = 15/π variable.
- Go back and make sure A = 30r - πr^2 that you have solved = 30(15/π) - π(15/π)^2 the problem. = 30(15/π) - π(225/π^2 ) = 450/π - 225/π = 225/π
The maximum area of the window is 225/π square feet.
15 - πr
Revised: Spring 2004 Source: Learning Skills Center The University of Texas at Austin Student Learning Assistance Center (SLAC) Texas State University-San Marcos
