Geometrical application of calculus question (1 Viewer)

[redorange]

This question has got me stuck..

I've done part a) but stuck on b)

Would be great if you guys could help and show me how you did it!

Thanks!

[answer is 6.25m^2]


edit: oh crap... this is a 2u question, not 3u. Can a mod move it if its necesary?

 

[pLuvia]

Find the first derivative, find the stat point, then using 2nd derivative or using the 1st derivative just show it is a maximum turning point

 

[abcd9146]

okay.. this may take a while... working from the computer
PART A

x²+y²=25
y²=25-x²
y=(25-x²)^1/2

A=xy/2
sub y into that
A=(1/2)x(25-x²)^1/2

which is part A done

PART B
A=(x/2)(25-x²)^1/2

A'=(1/2)(25-x²)^1/2+(1/2)(25-x²)^-1/2(-2x)(x/2)
A'=(1/2)(25-x²)^1/2+(-x²)/2(25-x²)^1/2
A'=(25-x²-x²)/[2(25-x²)^1/2]
A'=(25-2x²)/[2(25-x²)^1/2]

Stat Points at A'=0
0=(25-2x²)/[2(25-x²)^1/2]
0=(25-2x²)
2x²=25
x²=25/2
x=5/2^1/2
x=5.2^1/2/2

prove its max area (i use a table, but you may also use the 2^nd derivative.
x - 5.2^1/2/2 +
y + 0 -

therefore max TP (ie area)

finally
A=(1/2)(5.2^1/2/2)[(25-(5.2^1/2/2)²)^1/2]
A=6.25m²

 

[redorange]

Ahh thanks alot.

The differentiating in part B is the one that confused me.. i hate differentiating with square roots