differentiation maximum area question Watch

ErniePicks
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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.
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Zacken
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(Original post by ErniePicks)
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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 x and y here as well. You'd label the two opposite sides x and the other two opposite sides y and maximise the area under the constraint that 12x + 2y = 2000 \iff y = 1000 - 6x. Can you take it from here?
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ghostwalker
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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.
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genius10
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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...
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Zacken
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(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.
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genius10
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(Original post by Zacken)
Look at my answer, yours isn't quite right.
corrected
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Zacken
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(Original post by genius10)
corrected
Looks good. :borat:
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ErniePicks
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(Original post by Zacken)
Should be just x and y here as well. You'd label the two opposite sides x and the other two opposite sides y and maximise the area under the constraint that 12x + 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
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Zacken
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(Original post by ErniePicks)
Yeah sorry i thought it was 3 dimensions which is what was confusing me
Awesome.
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