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# differentiation maximum area question watch

1. 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.
2. (Original post by ErniePicks)

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 and here as well. You'd label the two opposite sides and the other two opposite sides and maximise the area under the constraint that . Can you take it from here?
3. 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.
4. Classic question

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...
5. (Original post by genius10)
Classic question

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.
6. (Original post by Zacken)
Look at my answer, yours isn't quite right.
corrected
7. (Original post by genius10)
corrected
Looks good.
8. (Original post by Zacken)
Should be just and here as well. You'd label the two opposite sides and the other two opposite sides and maximise the area under the constraint that . Can you take it from here?
Yeah sorry i thought it was 3 dimensions which is what was confusing me
9. (Original post by ErniePicks)
Yeah sorry i thought it was 3 dimensions which is what was confusing me
Awesome.

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