# a level differentiation

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#1
i'm not sure how to approach this question.
any help would be appreciated!
Last edited by abovethecl0uds; 1 month ago
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1 month ago
#2
(Original post by abovethecl0uds)
i'm not sure how to approach this question.
any help would be appreciated!
The area obviously depends on x and y, and you should be able to get an equation for how x and y are related.
So get the area as a function of x and differentiate to find the maximum.

Note there may be simpler ways of doing this, but youve posed it as a calculus problem.
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#3
(Original post by mqb2766)
The area obviously depends on x and y, and you should be able to get an equation for how x and y are related.
So get the area as a function of x and differentiate to find the maximum.

Note there may be simpler ways of doing this, but youve posed it as a calculus problem.
oh i see!
this is what i've done. i'd be grateful if you could check if it's correct
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1 month ago
#4
(Original post by abovethecl0uds)
oh i see!
this is what i've done. i'd be grateful if you could check if it's correct
Id guess the answer is a square, so that looks ok.

Note to keep the numbers a bit simpler, by symmetry, you could just have considered the problem in one quadrant.

Also, Id leave the pythagoras relationship as a quadratic and possibly maximize A^2. Would make the numbers a bit easier.
Last edited by mqb2766; 1 month ago
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#5
(Original post by mqb2766)
Id guess the answer is a square, so that looks ok.

Note to keep the numbers a bit simpler, by symmetry, you could just have considered the problem in one quadrant.

Also, Id leave the pythagoras relationship as a quadratic and possibly maximize A^2. Would make the numbers a bit easier.
that's so smart! i didn't realise the area of the rectangle in the circle must be at its maximum when all sides are equal in length eg) when it's a square.

i'm not quite sure what you mean by this though?
'Also, Id leave the pythagoras relationship as a quadratic and possibly maximize A^2.'
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1 month ago
#6
(Original post by abovethecl0uds)
that's so smart! i didn't realise the area of the rectangle in the circle must be at its maximum when all sides are equal in length eg) when it's a square.

i'm not quite sure what you mean by this though?
'Also, Id leave the pythagoras relationship as a quadratic and possibly maximize A^2.'
In the positive quadrant so a 1/4 of the original rectangle
A^2 = x^2(100-x^2)
Which is trivially maximized when x^2=50 (even without calculus) as its 1/2 way between the 2 roots of the quadratic in x^2.
Last edited by mqb2766; 1 month ago
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#7
(Original post by mqb2766)
In the positive quadrant so a 1/4 of the original rectangle
A^2 = x^2(100-x^2)
Which is trivially maximized when x^2=50 (even without calculus) as its 1/2 way between the 2 roots of the quadratic in x^2.
ah i see thanks so much for explaining!
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