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secondary 4 | A Maths
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Good evening! I need help with just part ii will do :)
Date Posted: 4 years ago
Views: 200
Hi Candice! I will look at it again at 3 am.
Oh its fine then! U shld rest at 3am haha its not urgent anyways :)
A = ½x √(441 - 42x) = ½x(441 - 42x)^½
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)
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)
Thank u so much
Welcome. If any doubts or need clarification on the working feel free to ask.
Hi Candice! There is an inaccuracy in the given answer for part iv, since the area cannot be zero.
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done
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Hi Candice! Here are my workings for the question. I disagree with the inequality sign of the solution for part iv.
Date Posted:
4 years ago
Thank u so much for the help once again :)
The inequality is correct. You can fold the paper all the way such that x = 0. Therefore A = 0. Which just means no triangle is formed.
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.
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.
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