Question

of all rectangles that have a perimeter of 34 ft, find the dimension with the largest area? What is its area?

Answer 1

calculus or algebra?

Answer 2

Because we could go either route here.

Answer 3

calculus

Answer 4

First all the perimeter is given to you.
We have 2L+2W=34
We want to maximize the area, A=LW

Answer 5

So solve 2L+2W=34 for either L or W.
Doesn't matter which one you choose.

Answer 6

how about L

Answer 7

Your current objective is to write A in terms of one variable (be it L or W but not both)

Answer 8

So what do you get when you solve 2L+2W=34

Answer 9

for L

Answer 10

L=17-W?

Answer 11

So recall A=LW
so we now have A=(17-W)W
Distribute there
then find A'

Answer 12

A' being dA/dW

Answer 13

why is there a W on the outside

Answer 14

equation

Answer 15

A=L*W

Answer 16

L=(17-W)

Answer 17

so A=(17-W)*W

Answer 18

I replaced L in A=LW with (17-W)

Answer 19

A=17-2W ?

Answer 20

That is A' or A?

Answer 21

I think that you mean A'=17-2W

Answer 22

A'

Answer 23

Now find your critical numbers
by finding when A'=0

Answer 24

17-2W=0 when W=?

Answer 25

W=8.5

Answer 26

Now you can find L
These will be the dimensions that maximize the area
and therefore you can also find the maximum area
you can verify this is a max by finding A''
A''=-2<0 therefore W=8.5 is a max

Answer 27

well W=8.5 is where the max occurs *

Answer 28

So am i suppose to take plug in the W into the original equation or take the second derivative of the equation A'=17-2W?

Answer 29

so i plugged 8.5 back into the original equation and L= 8.5 as well. That doesn't really make sense because it's a rectangle not a square

Answer 30

You need find L for when W=8.5
remember the relationship between W and L was
L=17-W

Answer 31

Right L=8.5

Answer 32

rectangles are squares when their side lengths are the same

Answer 33

And a square is what gives a max area

Answer 34

So the actual max area is 8.5*8.5

Answer 35

I'm still not following. How is a rectangle same as a square?

Answer 36

a rectangle is a square when the lengths of a rectangle are the same

Answer 37

therefore a rectangle can be a square

Answer 38

but not all rectangles are squares

Answer 39

so some rectangles are squares

Answer 40

okay so the area is 72.25

Answer 41

anymore questions?

Answer 42

I can also show you algebraically you will get the same dimensions for your rectangle.

Answer 43

Yes i have another question

Answer 44

\[2L+2W=34 \\ L+W=17 \\ A=LW=(17-W)W \\A=-W^2+17W \\ \text{ A is a parabola open downward, therefore it has a max and not a min}\\ A=-(W^2-17W) \\A=-(W^2-17W+(\frac{17}{2})^2)+(\frac{17}{2})^2 \\ A=-(W-\frac{17}{2})^2+(\frac{17}{2})^2 \\ \text{ the vertex where the max occurs is } (\frac{17}{2}, (\frac{17}{2})^2) \text{ So we have a \max area when } \\ W=\frac{ 17}{2} \text{ and } A=(\frac{17}{2})^2\]
And so L=17/2

Answer 45

so all squares are rectangles but not all rectangles are squares

Answer 46

what is the other question

Answer 47

From a thin piece of cardboard 20 in by 20 in, square corners are cut out so that the sides can be folded up to make a box. What dimensions will yield a box of maximum volume? What is the maximum volume?

Answer 48

Ok a drawing would help here

Answer 49

Can you draw it?

Answer 50

I'm sorry but there is no picture provided in my math textbook :(

Answer 51

Act like you are constructing an open box from some cardboard

Answer 52

And construct it exactly like it says

Answer 53

No lol I was asking you to draw the picture

Answer 54

I can draw the picture
The construction of the picture is for you