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AS Differentiation question please help

The angle C of triangle ABC is always right angle.
a) If the sum of CA and CB is 6cm , find the maximum area of triangle
b) If, on the other hand, the hypotenuse is to be kept equal to 4cm and the sides CA, CB allowed varying, finding the maximum area of the triangle.

i have done part a but i am stuck on part b

please help me with part b. I have attached my workings so far


thank you
P_20151007_051022.jpg P_20151007_051022.jpg
(edited 8 years ago)
Reply 1
BUMP
Please help me on part b
Do you have the other angles? I'm not sure it's doable without them if you only have two bits of data (one side, one angle)
Original post by bigmansouf
BUMP
Please help me on part b


I can't find a way of doing this with only AS knowledge. Can find two ways of doing it with A2 knowledge but maybe I'm just having a thick moment.
Original post by bigmansouf
BUMP
Please help me on part b


I found a way of doing it with only AS knowledge. Other than differentiation, do you know any ways of finding minimum/maximum values of functions?
Reply 5
Original post by morgan8002
I found a way of doing it with only AS knowledge. Other than differentiation, do you know any ways of finding minimum/maximum values of functions?

please can you share the way you did it for part b
Reply 6
Original post by morgan8002
I found a way of doing it with only AS knowledge. Other than differentiation, do you know any ways of finding minimum/maximum values of functions?

only completing the square comes to mind
Original post by bigmansouf
only completing the square comes to mind


Can you do any operations on the last line of your working to turn the problem into maximisation of a quadratic?
Reply 8
Original post by morgan8002
Can you do any operations on the last line of your working to turn the problem into maximisation of a quadratic?


thank you for the hint i was able to find that b=h= 22 2 \sqrt{2}

A= 4cm square
Original post by bigmansouf
thank you for the hint i was able to find that b=h= 22 2 \sqrt{2}

A= 4cm square


You're welcome.

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