• I'm having difficulty solving this problem to minimize cost of a rectangular box, the problem states:. A rectangular box holds a fixed volume of V cubic feet. The material used for the sides of the box cost A dollars per square foot, and the material for the top and bottom of the box costs B dollars per square foot.

What are the dimensions of thi type of box that will have a minimum cost for material for the box? In terms of the parameters V, A, and B.

Thanks for the help, but i just have one quick question, how did B get substituted out? Going from:. Therefore your box has no minimal cost. You can always make the box cost less by making it narrower in the direction of the more costly material. As you use less and less of the more expensive material, such a box come to resemble long tube, which might not fit what you are packing. Therefore you would typically need a second size constraint in order to create an objective function with a true minimum. Your situation has been kicked up some notches. First symbolize your field as extra than a number of things with linear dimensions and incorporate the welds.

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There are 4 welds connecting the backside to the factors and four welds on the corners. If the field is of uniform top and has an oblong ingredient whilst seen from above, the ingredient welds are all a similar length. If it resembles an oblong prism open at precise opposite pairs of backside welds are a similar length. There are 2 factors with area xz and a pair of with area yz.

Throw on the value of the lid which will fall out of each partial spinoff expression and commence diagnosis.

I'm having difficulty solving this problem to minimize cost of a rectangular box, the problem states: A rectangular box holds a fixed volume of V cubic feet.

Update: Thanks for the help, but i just have one quick question, how did B get substituted out? Answer Save. Michael T Lv 5. B did not get substituted out. It got inadvertently deleted. It's back now. Props for making an attempt yourself instead of just asking us for the answer. Still have questions? Get your answers by asking now.Minimizing Costs A pencil cup with a capacity of 36 in.

The shape of the cup is a rectangle box with square base and open top. Assume that the length, width and height of the pencil cup is lland h respectively. The amount of material used to make the cup is given by the total surface area of the open cuboid. Substitute 36 l 2 for h in the above formula. Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees! Subscribe Sign in. Operations Management. Chemical Engineering.

## Find dimensions to minimize cost of box

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### What dimensions minimize the cost?

Social Science. Applied Calculus for the Manageria Problem 1CQ. Problem 2CQ. Problem 1E. Problem 2E. Problem 3E.

Problem 4E.Using given information about the Volume, express the height h as a function of the width w. Take the derivative of the Cost with respect to width and set it to zero to determine critical point s. How do you find the dimensions of the box that minimize the total cost of materials used if a rectangular milk carton box of width w, length l, and height h holds cubic cm of milk and the sides of the box cost 4 cents per square cm and the top and bottom cost 8 cents per square cm?

Alan P. Mar 11, Write an expression for the Cost in terms of only the width w. Related questions How do you find two numbers whose difference is and whose product is a maximum? How do you find the dimensions of a rectangle whose area is square meters and whose How do you find the dimensions of the rectangle with largest area that can be inscribed in a Question b1. The fencing for the north and south How do you find the volume of the largest right circular cone that can be inscribed in a sphere How do you find the dimensions of a rectangular box that has the largest volume and surface area What are the dimensions of a box that will use the minimum amount of materials, if the firm How do you find the dimensions that minimize the amount of cardboard used if a cardboard box See all questions in Solving Optimization Problems.

Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up. Suppose, to build a box a rectangular solid of fixed volume and square base, the cost per square inch of the base and top is twice that of the four sides. You are off to a good start. Note that the actual cost plays no role in determining where the minimum occurs.

It will change the value of that minimum, but not the location. I expect that you should be able to finish the problem from here. Sign up to join this community. The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. Minimize the cost of a box of fixed volume if the sides are twice as expensive as the base and top Ask Question.

Asked 1 year, 8 months ago. Active 1 year, 7 months ago. Viewed times. Mack A. Mack 33 5 5 bronze badges. Active Oldest Votes. Xander Henderson Xander Henderson I had to look up the AM-GM inequality. The Overflow Blog.

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Socializing with co-workers while social distancing. Featured on Meta. Community and Moderator guidelines for escalating issues via new response…. Feedback on Q2 Community Roadmap. Question to the community on a problem.In this section we are going to look at optimization problems. In optimization problems we are looking for the largest value or the smallest value that a function can take. We saw how to solve one kind of optimization problem in the Absolute Extrema section where we found the largest and smallest value that a function would take on an interval.

In this section we are going to look at another type of optimization problem. Here we will be looking for the largest or smallest value of a function subject to some kind of constraint. The constraint will be some condition that can usually be described by some equation that must absolutely, positively be true no matter what our solution is. This section is generally one of the more difficult for students taking a Calculus course.

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One of the main reasons for this is that a subtle change of wording can completely change the problem. The first step in all of these problems should be to very carefully read the problem. In identifying the constraint remember that the constraint is the quantity that must be true regardless of the solution. It is however easy to confuse the two if you just skim the problem so make sure you carefully read the problem first! In all of these problems we will have two functions. The first is the function that we are actually trying to optimize and the second will be the constraint. In this problem we want to maximize the area of a field and we know that will use ft of fencing material.

So, the area will be the function we are trying to optimize and the amount of fencing is the constraint. The two equations for these are.

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However, if we solve the constraint for one of the two variables we can substitute this into the area and we will then have a function of a single variable. Having these limits will also mean that we can use the process we discussed in the Finding Absolute Extrema section earlier in the chapter to find the maximum value of the area. That means our only option will be the critical points.

So according to the method from Absolute Extrema section this must be the largest possible area, since the area at either endpoint is zero.

The dimensions of the field that will give the largest area, subject to the fact that we used exactly ft of fencing material, are x In the previous problem we used the method from the Finding Absolute Extrema section to find the maximum value of the function we wanted to optimize. Also, even if we can find the endpoints we will see that sometimes dealing with the endpoints may not be easy either.

Method 1 : Use the method used in Finding Absolute Extrema. This is the method used in the first example above. If these conditions are met then we know that the optimal value, either the maximum or minimum depending on the problem, will occur at either the endpoints of the range or at a critical point that is inside the range of possible solutions.

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There are two main issues that will often prevent this method from being used however. First, not every problem will actually have a range of possible solutions that have finite endpoints at both ends.Sam wants to build a garden fence to protect a rectangular square-foot planting area. His next-door neighbor agrees to pay for half of the fence that borders her property; Sam will pay the rest of the cost.

Link will open in a new tab. Your first job is to develop a function that represents the quantity you want to optimize. It can depend on only one variable.

The steps:. Draw a picture of the physical situation. See the figure. Write an equation that relates the quantity you want to optimize in terms of the relevant variables.

Remember that Sam is only paying for half of the cost of the right side of the fence, since his neighbor is paying for that other half. If necessary, use other given information to rewrite your equation in terms of a single variable. The cost C currently depends on two variables, y and x. We must then eliminate y as a variable. One choice would be to closely examine the graph to determine the value of x that minimizes C. Take the derivative of your equation with respect to your single variable.

Then find the critical points. Justify your maxima or minima either by reasoning about the physical situation, or with the first derivative test, or with the second derivative test. Hence this single critical point gives us a minimum for the cost C :. Determine the maxima and minima as necessary.

Remember to check the endpoints if there are any. Return to Optimization Problems. Want access to all of our Calculus problems and solutions? Head over to our partners at Chegg Study and gain 1 immediate access to step-by-step solutions to most textbook problems, probably including yours; 2 answers from a math expert about specific questions you have; AND 3 30 minutes of free online tutoring.

Please visit Chegg Study now. Log in. Calculus Optimization Problem: What dimensions minimize the cost of a garden fence? Stage I: Develop the function. The steps: 1. Stage II: Maximize or minimize the function.

Do you need immediate help with a particular textbook problem? I agree to the Terms and Privacy Policy. The comment form collects the name and email you enter, and the content, to allow us keep track of the comments placed on the website.In Section 3. In this section, we apply the concepts of extreme values to solve "word problems," i.

We start with a classic example which is followed by a discussion of the topic of optimization. A man has feet of fencing, a large yard, and a small dog. He wants to create a rectangular enclosure for his dog with the fencing that provides the maximal area.

Optimization Calculus - Fence Problems, Cylinder, Volume of Box, Minimum Distance & Norman Window

What dimensions provide the maximal area? One can likely guess the correct answer -- that is great. We will proceed to show how calculus can provide this answer in a context that proves this answer is correct. It helps to make a sketch of the situation. Either way, drawing a rectangle forces us to realize that we need to know the dimensions of this rectangle so we can create an area function -- after all, we are trying to maximize the area. We do not yet know how to handle functions with 2 variables; we need to reduce this down to a single variable.

We know more about the situation: the man has feet of fencing. By knowing the perimeter of the rectangle must bewe can create another equation:.

## What dimensions minimize the cost of a rectangular box?

This is the maximum. Thus the dimensions of the rectangular enclosure with perimeter of ft. This example is very simplistic and a bit contrived.

After all, most people create a design then buy fencing to meet their needs, and not buy fencing and plan later. But it models well the necessary process: create equations that describe a situation, reduce an equation to a single variable, then find the needed extreme value.

The equations are often not reducible to a single variable hence multi--variable calculus is needed and the equations themselves may be difficult to form.

Understanding the principles here will provide a good foundation for the mathematics you will likely encounter later.