Solving for perimeter and dimensions of a rectangle

In summary: That is the derivative of P=2L+32/L which is P= 2L+32/LNow, when is that 0?P=2+ 32/L^2P'= -64/L^3Set that equal to 0.-64/L^3= 0-64= 0L= 0That is a critical point. But you cannot have a rectangle with area 16 if one side is 0! So that is not a solution.The only other possibility is that L= 0. In summary, the conversation is about finding the smallest perimeter for a rectangle with an area of 16. The solution involves finding
  • #1
646
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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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  • #2
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?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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?
 
  • #6
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
 

What is the formula for finding the perimeter of a rectangle?

The formula for finding the perimeter of a rectangle is P = 2(l + w), where P is the perimeter, l is the length, and w is the width.

How do you find the dimensions of a rectangle given its perimeter?

To find the dimensions of a rectangle given its perimeter, you can use the formula P = 2(l + w) and plug in the value of the perimeter. Then, solve for either the length or the width by rearranging the equation.

Can you solve for the perimeter of a rectangle if you only know one side?

No, the perimeter of a rectangle cannot be determined if you only know one side. You need to know at least two sides in order to solve for the perimeter.

What is the difference between perimeter and area of a rectangle?

Perimeter is the distance around the outside of a shape, while area is the measure of the space inside the shape. The area of a rectangle is found by multiplying the length by the width, while the perimeter is found by adding the lengths of all sides.

How do you find the dimensions of a rectangle with a given area?

To find the dimensions of a rectangle with a given area, you can use the formula A = l * w, where A is the area, l is the length, and w is the width. Plug in the area and solve for either the length or the width by rearranging the equation.

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