Question

WILL MEDAL

Optimization: If you have 60 meters of fencing and want to enclose a rectangular area up against a long, straight wall, what is the largest area you can enclose?

Answer 1

Thus, w=70

Answer 2

Have you studied calculus? Optimisation problems?

Answer 3

Yes! We are just learning Optimization now. I just dont know how to start it

Answer 4

We can say A=lw, but one side already has the wall, so we do not need fencing for one of those 4 sides

Answer 5

Ok then we need to find the maximum area of rectangle
A = lw where l = length and w = width

so the 3 sides = 60 so we have the equation
l + 2w = 60

ok so far?

Answer 6

the area = lw

Answer 7

oh okay i get it!

Answer 8

so then how do we estimate the values of the length and width from the 60meters?

Answer 9

now we need to find an expression for A in terms of one of the variables:-

2w + l = 60
l = 60 - 2w
Substituting in the equation A = lw;-

A = w(60 - 2w)

OK/

Answer 10

yeah that makes sense.

Answer 11

should i distribute the w?

Answer 12

and then get the derivative?

Answer 13

yes

then you find the derivative dA/dw and equate it to zero
Then solve for w

Answer 14

knowing the value of w you use l + 2w = 60 to find the length of the rectangle.
Then of course the area required is l * w.

Answer 15

so: 60w-2w^2

derivative= 60-4x

Answer 16

i mean 4w

Answer 17

60-4w=0

-4/60 = w

Answer 18

w = 60 / 4

Answer 19

oh right XD

Answer 20

W= 15 so L= 45

Answer 21

60 - 4w = 0
4w = 60

remember to put the derivative = zero.

Answer 22

So Area=675?

Answer 23

60-4x=0

Answer 24

L + 2(15) = 60
L = 60 - 2(15) = 60 - 30

Answer 25

take your time..

Answer 26

oh right i forgot to plug them back into the equation

Answer 27

so L=30 and W=15

Answer 28

And so area=450

Answer 29

yeah

Answer 30

This makes much more sense XD

Answer 31

Thank you so much!

Answer 32

yw

Answer 33

and units would be meters?