Question

Optimization Problem, (Question on a person's example).

Answer 1

Original Question: Find the dimensions of a rectangle with area 343 m2 whose perimeter is as small as possible. (If both values are the same number, enter it into both blanks.)

Answer 2

\[P(x)=2x+\frac{686}{x} \\ P(\sqrt{343})=2 \sqrt{343}+\frac{686}{\sqrt{343}} \\ P(\sqrt{343})=2 \sqrt{343}+\frac{686 \sqrt{343}}{343} \\ P(\sqrt{343})=2\sqrt{343}+2 \sqrt{343} \\ P(\sqrt{343})=4 \sqrt{343}\]

Answer 3

and also the major hint where they plugged into is standing right next to the (sqrt(343))
which is P

Answer 4

I don't see 7 sqrt(343) anywhere

Answer 5

ops, i meant 7sqrt(7)

Answer 6

\[\text{ is } 7^2 \cdot 7=343 ?\]

Answer 7

\[\sqrt{343}=\sqrt{7^2 \cdot 7}=\sqrt{7^2} \sqrt{7}=7 \sqrt{7}\]

Answer 8

what happened to 4sqrt(343)?

Answer 9

How does that factor into getting the answer?

Answer 10

\[P(\sqrt{343})=4\sqrt{343} \\ \text{ smallest perimeter is } 4 \sqrt{343} \\ \text{ and } x =\sqrt{343} \text{ gave us that value } \\ \text{ and then you find the other dimension which was defined as } y \]

Answer 11

P was to thing we minimized with P'=0

Answer 12

P'' was to double check the value we got when P'=0 was indeed a min

Answer 13

What is the other dimension that was defined by y?

Answer 14

I get up to the part of how x=sqrt(343)

Answer 15

you have a rectangle
the dimensions were given in the first line they they were x and y...

Answer 16

you find those values later on in the problem

Answer 17

In this example, small and large value have the same answer as sqrt(343).

Answer 18

If x is a critical number in the domain of the original function
and if P''(x)>0 then you have a min at x
and if P''(x)<0 then you have a max at x

Answer 19

why do you call it small and large value?

Answer 20

In this example of a practice version with the solution, it wants you to find the small and large value. (if both of the answers are the same value, then it counts for small and large)

Answer 21

what is the question though about the dimensions ?

Answer 22

the whole question is in picture 2 above your recent reply.

Answer 23

"where did the person plug sqrt(343) in?"

Answer 24

That question?

Answer 25

I thought you had another question. Sorry.

Answer 26

Oh, sorry for the confusion. I see where they plugged in sqrt(343) in. I'll open a new question for the 2nd question.

Answer 27

you can ask it here
I just trying to find out the other question

Answer 28

like if you had a question about the dimensions

Answer 29

like you mentioned something about the smaller and larger one
I didn't know if there was a question in there about smaller and larger one

Answer 30

I see how they plugged in sqrt(343) into P(x) to get 4sqrt(343) the smallest number. So I got it now, but how did they figure out the other value? The end of the solution could use a bit more clarification for me.

Answer 31

ok do you see that y=343/x ?

Answer 32

this was because xy=343

Answer 33

yea

Answer 34

you can find y by replacing x with sqrt(343)

Answer 35

oh, so thats how they gotten sqrt(343) for y value

Answer 36

\[y=\frac{343}{\sqrt{343}} \\ y=\frac{343}{\sqrt{343}} \cdot \frac{\sqrt{343}}{\sqrt{343}} \\ \\ y=\frac{343 \sqrt{343}}{343} =\sqrt{343}\]

Answer 37

and 7sqrt(7) is simply for "looks?"

Answer 38

in some algebra classes they called that the simplified version

Answer 39

unless somewhere prefers it to be simplified

Answer 40

yea

Answer 41

so yep for looks

Answer 42

okay, I understand the problem now completely. Thanks for bearing with me :)

Answer 43

np