Solving for perimeter and dimensions of a rectangle

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Homework Help Overview

The discussion revolves around finding the smallest perimeter of a rectangle given a fixed area of 16. The problem involves applying calculus concepts to optimize the perimeter function based on the relationship between length and width.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formulation of the perimeter equation in terms of a single variable and the application of calculus to find critical points. Questions arise about the interpretation of derivatives and their role in identifying minimum values.

Discussion Status

Participants are actively engaging with the calculus concepts needed to solve the problem. Some guidance has been offered regarding finding critical points and evaluating derivatives, but there is no explicit consensus on the next steps or the application of these concepts.

Contextual Notes

There is a noted difficulty in applying calculus to the problem, and participants are questioning the clarity of the equations presented. The original poster expresses uncertainty about the process, indicating a potential gap in understanding how to connect the mathematical formulation with the problem context.

Lancelot59
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I understand calculus, I just don't understand how it is applied to solve these sorts of problems.

Homework Statement


What is the smallest perimeter for a rectangle with an area of 16, and what are its dimensions?

Homework Equations


A=L*W
P=2W+2L

The Attempt at a Solution



I managed to get the perimeter equation down to one variable like so:

Area is 16.

A=LW
16=LW
16/L=W

P=2L+2(16/L)
P=(2L2+32)/L

and this is where I stop. I just don't see how this can be applied to the situation stated.
 
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Well you now have a function, P, of just one variable, L. You want to find when this function is a minimum? Have you been shown how to find the minimum points of a function using calculus?
 
Yes I have.

I need to look for locations where the first derivative is zero and the second derivative is positive on both sides of the critical point.

I'm just having a tricky time actually applying calculus to these situations.
 
Ok. The easiest way to do these types of problems is probably to first find all of the critical point of the function (i.e. where the derivative is zero or undefined). From there, you can then check whether the critical point(s) you find are local maximums/minimums or inflections by evaluating the second derivative at those points.
 
Lancelot59 said:
I understand calculus, I just don't understand how it is applied to solve these sorts of problems.

Homework Statement


What is the smallest perimeter for a rectangle with an area of 16, and what are its dimensions?


Homework Equations


A=L*W
P=2W+2L

The Attempt at a Solution



I managed to get the perimeter equation down to one variable like so:

Area is 16.

A=LW
16=LW
16/L=W

P=2L+2(16/L)
P=(2L2+32)/L

and this is where I stop. I just don't see how this can be applied to the situation stated.

Lancelot59 said:
Yes I have.

I need to look for locations where the first derivative is zero and the second derivative is positive on both sides of the critical point.

I'm just having a tricky time actually applying calculus to these situations.
Perhaps because you are not trying! What is the derivative of P= 2L+ 32/L? Where is that derivative 0?
 
P=(2L^2+32)/L

Where do you get the power of 2?

"P=2L+2(16/L)
P=(2L2+32)/L" ??

P=2L+2(16/L)
P=2(L+16/L)

or

2L+32/L
L+16/L
 

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