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differentiation maximum area question

mathseee.jpg
Just no clue how to do this i've spent ages. I don't know what to label the sides, suaully it's just x and y but on this one i dont know. Just please any help in labelling what the different sides would be and i think can do it.
Reply 1
Avatar for Zacken
Zacken
22
Original post by ErniePicks
mathseee.jpg
Just no clue how to do this i've spent ages. I don't know what to label the sides, suaully it's just x and y but on this one i dont know. Just please any help in labelling what the different sides would be and i think can do it.


Should be just xx and yy here as well. You'd label the two opposite sides xx and the other two opposite sides yy and maximise the area under the constraint that 12x+2y=2000  ⟺  y=1000−6x12x + 2y = 2000 \iff y = 1000 - 6x. Can you take it from here?
May be worth commenting that the farmer will use the full £2000 (hence the =2000). If she didn't, then she could increase either/both of the sides to get a greater area.
Reply 3
Avatar for genius10
genius10
7
Classic question :biggrin:

Set up two equations for the side lengths, a and b, and the area
(sorry no latex idk how to use it)

12a + 2b = 2000 and area=a*b

not substitute b into the area equation to get an equation linking area with a

area=a(1000-6a)=1000a-6a^2

Firstly, you know that this equation is correct up to this point because when a=0 the area is or when a=1000/6 (when b=0)

Now find the turning point of the quadratic, (this is where the area is at a local maximum).

should be a piece of cake to solve from there...
(edited 7 years ago)
Reply 4
Avatar for Zacken
Zacken
22
Original post by genius10
Classic question :biggrin:

Set up two equations for the side lengths, a and b, and the area
(sorry no latex idk how to use it)

6a + b = 2000 and area=a*b

not substitute b into the area equation to get an equation linking area with a

area=a(2000-6a)=2000a-6a^2

Firstly, you know that this equation is correct up to this point because when a=0 the area is or when a=2000/6 (when b=0)

Now find the turning point of the quadratic, (this is where the area is at a local maximum).

should be a piece of cake to solve from there...


Look at my answer, yours isn't quite right.
Reply 5
Avatar for genius10
genius10
7
Original post by Zacken
Look at my answer, yours isn't quite right.


corrected
Reply 6
Avatar for Zacken
Zacken
22
Original post by genius10
corrected


Looks good. :borat:
Reply 7
Avatar for ErniePicks
ErniePicks
OP
2
Original post by Zacken
Should be just xx and yy here as well. You'd label the two opposite sides xx and the other two opposite sides yy and maximise the area under the constraint that 12x+2y=2000  ⟺  y=1000−6x12x + 2y = 2000 \iff y = 1000 - 6x. Can you take it from here?


Yeah sorry i thought it was 3 dimensions which is what was confusing me
Reply 8
Avatar for Zacken
Zacken
22
Original post by ErniePicks
Yeah sorry i thought it was 3 dimensions which is what was confusing me


Awesome.

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