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secondary 4 | A Maths

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Good evening! I need help with just part ii will do :)

use chain rule,

dA/dx

= ½(441 - 42x)^½ + ½x(½)(-42)(441 - 42x)^(-½)

= ½(441 - 42x)^½ - 21x/(2(441 - 42x)^(-½) )

= ½√(441 - 42x) - 21x/(2√(441 - 42x) )

when dA/dx = 0,

½√(441 - 42x) - 21x/(2√(441 - 42x)) = 0

½√(441 - 42x) = 21x/(2√(441 - 42x))

√(441 - 42x) = 21x/(√(441 - 42x))

441 - 42x = 21x

63x = 441

x = 7

A = ½(7) √(441 - 42(7))

A = 7/2 √147

≈ 42.4352448

= 42.4 (3.s.f)

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Likewise, if you fold such that x = 10.5, you've just folded the paper into half and no triangle is formed.

For both cases, this means PR = QR and P = Q.

Both cases satisfy the notion of the corner touching the opposite sides of the paper just that there's no triangle.

What's for sure is that x cannot be negative. In the actual graph, values for A exist for x < 0. A is negative for those x values. Just that we do not consider them since area can never be negative.