A very easy question that im stuck with (1 Viewer)

[kooltrainer]

The method i used (used graph B)
Area = xy
= x root. (4-x^2) [subbing y = root. (4-x^2) ]
let derivative = 0

and u'll get x = root 2
and u'll get y = root (3.5)


The second method i used (used graph A )
Area = 2xy
= 2x root.(4-x^2) [ subbing y=root (4-x^2) ]

let derivative = 0

and u'll get x = root2
and u'll get y = root 2
so dimension to 2root2 and root 2



similar method, different answer? apparently, method 2 is correct.. why is method 1 wrong? / / ? ? ?

 

[powerdrive]

In the second method u used, graph A, what u are actually doing is letting there be a variable point on the curve, lets say P(x,y), meaning it is x units across and y units up from the origin. This is the correct method, and hence the base of the rectangle is 2x units.

With the first method in graph B, what's happened there is the whole base of the rectangle is only x units, meaning the variable point would be seen as P(1/2x,y). This is wrong as the point must be P(x,y) to be variable.

 

[kooltrainer]

noo, p(x/2 , y) can still be variable.. and graph b is perfectly normal..

 

[powerdrive]

lol sorry, had a feeling i was wrong after writing that, its been a while since i finished hsc

 

[Mark576]

(B) doesn't give the correct answer as you're dealing with a semi-circle, so you have to take special care here. It's hard to explain but essentially, the X in the diagram doesn't correspond to the x in the equation of the semi-circle, so you're obviously not going to get the correct answer. The x is defined in the equation as an ordinate, whereas the X in the diagram simply refers to a length.

 

[Steth0scope]

I don't think either of your methods at right.

You are subbing in the equation of the circle for Y into the Area equation. Instead of finding the area you'll be finding points of intersection (i think) which is why u got those funny figures in the first part.

I sorta just did it like this (hope u can follow on).

From the formula of the circle u knw that the radius is 2. This radius is also the diagonal of a square (half of rectange) formed when you try to inscribe the rectangle. Using pythagors therom (with the diagonal of a square being 2), you get each side with a length of root 2.

The base of the whole rectangle will be twice that - therefore two root 2 ?

Therefore dimensions = root 2 x 2 root 2.

If i could draw it you would understand so easily. Try draw a rectangle and then draw in the diagonal from the centre of the semi circle to one of the corners. Hopefully you'll get it.

 

[Mark576]

We're to find points of intersection that give the largest rectangle that can be inscribed in the semi-circle. So we find an expression for the area of the rectangle in terms of only one variable (x) and then solve dA/dx = 0 in order to find dimensions that will give the maximum area. This gives x = rt2 and y = rt2. Therefore the dimensions (using method (A)) will be 2rt2 and rt2, and the maximum area is 4 units^2. I've already explained why we must use method (A) as opposed to method (B).

 

[kooltrainer]

We're to find points of intersection that give the largest rectangle that can be inscribed in the semi-circle. So we find an expression for the area of the rectangle in terms of only one variable (x) and then solve dA/dx = 0 in order to find dimensions that will give the maximum area. This gives x = rt2 and y = rt2. Therefore the dimensions (using method (A)) will be 2rt2 and rt2, and the maximum area is 4 units^2. I've already explained why we must use method (A) as opposed to method (B).

i understand now a bit .. but one thing i don't get, why is the length (x) value from graph B different from length (2x) value from graph A? they're the same rectangle..
also, in graph B, u find out that X is rt2 .. u can divide that by 2 (that'll give u x cor ordinate) , then sub into semi-circle equation and find y value .. i dun see y is that wrong..

 

[Mark576]

But x =/= rt2, as the method used to derive an answer for x in graph (B) is incorrect. To answer your first question: the values are different because you can't use graph (B) to work out the maximum area because you have allowed the length to equal x, and you're assuming this x is equal to the x in the equation, allowing you to substitute and subsequently differentiate. But they're not equal, because the x in the equation refers to an ordinate, a certain distance from the origin whereas the x you used in the rectangle clearly doesn't. So it doesn't work out correctly. Does that help? Sorry, maybe I'm not being clear enough.

EDIT: And that's why you can use method (A), because the pronumeral used (x) in both equations represents the same thing.