Applied maximum and minimum problems problem

In summary: Now use the usual "find the minimum of a function" process.In summary, to minimize the cost of materials for the walls and front of a new branch bank with a floor area of 3500 ft^{2}, you need to minimize 2y+ 2.8x, where y= 3500/x and xy= 3500. This can be done by finding the minimum of the function of x alone, 2(3500/x)+ 2.8x, using the usual "find the minimum of a function" process.
  • #1
elitespart
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Optimization

1. A new branch bank is to have a floor area of 3500 ft[tex]^{2}[/tex]. It is to be a rectangle w/ 3 solid brick walls and a decorative glass front. The glass costs 1.8 times as much as the brick wall per linear foot. What dimensions of the building will minimize cost of materials for the walls and front?

I'm learning this stuff on my own and I need a little guidance on this problem. I know I have to write everything in terms of a single variable before I can derive. I'm having trouble understanding how to set up the problem. Am I looking to minimize surface area and does the volume come into play here? Thanks.

So I have y=3500/x. Do I have to use that in surface area or volume?
 
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  • #2


elitespart said:
1. A new branch bank is to have a floor area of 3500 ft[tex]^{2}[/tex]. It is to be a rectangle w/ 3 solid brick walls and a decorative glass front. The glass costs 1.8 times as much as the brick wall per linear foot. What dimensions of the building will minimize cost of materials for the walls and front?

I'm learning this stuff on my own and I need a little guidance on this problem. I know I have to write everything in terms of a single variable before I can derive. I'm having trouble understanding how to set up the problem. Am I looking to minimize surface area and does the volume come into play here? Thanks.

So I have y=3500/x. Do I have to use that in surface area or volume?
Neither, actually. Since you are not given any height you cannot find either the volume of the building or the area of the walls. And you do not want to miniize surface area: you are specifically asked to minimize the "cost".

You are given the cost in terms of "linear feet" so the cost of the wall will be proportional to its length. If x is the length of the front and back and y is the length of the sides, then we have 2 brick walls of length y and one brick wall of length x. Take the "cost" of one linear foot of brick wall to be "1". Then the total cost of brick walls will be 2y+ x. The remaining wall, of length x, is glass and its cost will be 1.8x. The total cost of the walls is 2y+ x+ 1.8x= 2y+ 2.8x. That is what you want to minimize subject to the condition that xy= 3500. Yes, y= 3500/x and you can replace y by that in 2y+ 2.8x to get a function of x only.
 

What is an applied maximum and minimum problem?

An applied maximum and minimum problem is a type of mathematical problem that involves finding the highest or lowest value of a given function within a specific range of values. These problems are commonly used in fields such as engineering, physics, and economics to optimize certain variables and make informed decisions.

What are some common examples of applied maximum and minimum problems?

Some common examples of applied maximum and minimum problems include finding the maximum profit for a company, determining the minimum amount of material needed to construct a bridge, and identifying the minimum amount of time required to complete a project.

What are the steps to solving an applied maximum and minimum problem?

The steps to solving an applied maximum and minimum problem include identifying the variables and constraints, setting up the appropriate equations or inequalities, finding the critical points by taking the derivative of the function, and using the critical points to determine the maximum or minimum value.

What are some strategies for solving applied maximum and minimum problems?

Some strategies for solving applied maximum and minimum problems include graphing the function to visualize the problem, using algebraic techniques such as substitution and factoring, and utilizing calculus methods such as taking derivatives and using optimization techniques.

Why are applied maximum and minimum problems important?

Applied maximum and minimum problems are important because they allow us to optimize variables and make informed decisions in various fields. They also help us understand the behavior of a function and its critical points, which can have practical applications in real-world scenarios.

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