Can anyone answer this question

[jtesttubes]

Draw a rectangle , start quadrant having one leg on the Axis -Y and other on the Axix -X and vertex on the curve
y=1-0.61x^2

You need to find the co-ordinates of the rectangle and find out which form of rectangle has got more area.

 

[tiny-tim]

Hi jtesttubes!

Show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[HallsofIvy]

Draw a rectangle , start quadrant having one leg on the Axis -Y and other on the Axix -X and vertex on the curve
y=1-0.61x^2

You need to find the co-ordinates of the rectangle and find out which form of rectangle has got more area.

Yes, I can answer that. Thank you for asking.

 

[jtesttubes]

Hi jtesttubes!

Show us what you've tried, and where you're stuck, and then we'll know how to help!

I have worked out a calculation, but i just want to confirm with you whether this is the way to work out this question, i got the co-ordinates of the rectangle. Is it correct

Area = x*y - basic formula for area.
A=x(1-0.61x^2)
A=x-0.61x^3
dA/dx = 1-1.83x^2
0=1-1.83x^2
x=sqrt(1/1.83)
x=0.739
y=1-0.61(0.739)^2
y=0.66667
coordinates are (0.739,0.6667)

 

[tiny-tim]

Hi jtesttubes!

(try using the X² tag just above the Reply box )

I have worked out a calculation, but i just want to confirm with you whether this is the way to work out this question, i got the co-ordinates of the rectangle. Is it correct

Area = x*y - basic formula for area.
A=x(1-0.61x^2)
A=x-0.61x^3
dA/dx = 1-1.83x^2
0=1-1.83x^2
x=sqrt(1/1.83)
x=0.739
y=1-0.61(0.739)^2
y=0.66667
coordinates are (0.739,0.6667)

yes, that looks fine.

(except you could have got y more quickly by noting that 0.61x² is 1/3 of 1.83x², and you know exactly what that is )

 

[jtesttubes]

Oh thanks i have some doubts on my workings now I am happy that it is the right way to work out, yes sometimes i do not follow shortcuts on working out this kind of equations.