why did the author took the square of area

[friar]

Q2: Find the dimension of a rectangle of maximum area that can be inscribed in a semicircle of radius 10cm. Hence find the maximum Area? Solution: let 2x and y be the length and with of the rectangle ABCD. Now in ∆OBC, by Pythagoras theorem, x2 + y2 = 102 ⟹ y2 = 100 – x2 ⟹ y = √(100 – x^2 ) Let A = square of Area i.e. A = (Length X width)2 A = (2x * y )2 A = 4x2y2 A = 4x2 (100 – x2) A = 400x2 – 4x4 -------------------------------(1) differentiating equation (1) wrt “x” so we get, dA/dx = 800x – 16x3.------------------------------(2) differentiating equation (2) wrt “x” so we get, (d^2 A)/〖dx〗^2 = 800 – 48x2. ---------------------------- (3) for critical points put dA/dx = 0. 800x – 16x3 = 0. 16 x (50 – x2) = 0. As x ≠ 0, so 50 – x2 = 0. ⟹ x2 = 50. ⟹ x = ±5√2 Put x = 5√2 in equation (2) so we get, (d^2 A)/〖dx〗^2 = 800 – 48(5√2)2. (d^2 A)/〖dx〗^2 = 800 – 2400= - 1600 (d^2 A)/〖dx〗^2 = -1600 < 0. So area is maximum at x = 5√2 ∴ Length = 2x = 10√2 width = y = 5√2 Area = 100. Answer. On 6th line why did the author took the square of area. Known Rules don’t allow that. regards

 

[HallsofIvy]

First, at least use ^ to indicate powers. (You did once!)

Q2: Find the dimension of a rectangle of maximum area that can be inscribed in a semicircle of radius 10cm. Hence find the maximum Area? Solution: let 2x and y be the length and with of the rectangle ABCD. Now in ∆OBC, by Pythagoras theorem, x^2 + y^2 = 10^2 ⟹ y2 = 100 – x^2 ⟹ y = √(100 – x^2 ) Let A = square of Area i.e. A = (Length X width)^2 A = (2x * y )^2 A = 4x^2y^2 A = 4x^2 (100 – x^2) A^2 = 400x^2 – 4x^4 -------------------------------(1) differentiating equation (1) wrt “x” so we get, dA/dx = 800x – 16x^3.------------------------------(2) differentiating equation (2) wrt “x” so we get, (d^2 A^2)/〖dx〗^2 = 800 – 48x^2. ---------------------------- (3) for critical points put dA/dx = 0. 800x – 16x3 = 0. 16 x (50 – x2) = 0. As x ≠ 0, so 50 – x2 = 0. ⟹ x2 = 50. ⟹ x = ±5√2 Put x = 5√2 in equation (2) so we get, (d^2 A)/〖dx〗^2 = 800 – 48(5√2)2. (d^2 A)/〖dx〗^2 = 800 – 2400= - 1600 (d^2 A)/〖dx〗^2 = -1600 < 0. So area is maximum at x = 5√2 ∴ Length = 2x = 10√2 width = y = 5√2 Area = 100. Answer. On 6th line why did the author took the square of area. Known Rules don’t allow that. regards

For simplicity. The derivative of A^2 is 2A dA/dx and since A is not 0, that will be 0 if and only if dA/dx is. So as far as finding x that maximizes A is concerned, it is the same as the x that maximizes A^2. And since A^2 does not have that square root, it is easier to work with.

 

[MarkFL]

You will find this a useful "shortcut" when optimizing distances as well.

 

[friar]

why did the author took the square of the area

Thanks for the answer. Waiting for more (if any). Wrote the above question in correct form. Just click the link below. http://www.yourfilelink.com/get.php?fid=828628 You need MS Word 2010 to view it. I think there is error in forum software. Appreciate if any member suggests some good mathematical Equa-tions writing programs. Regards. HallsofIvy For simplicity. The derivative of A^2 is 2A dA/dx and since A is not 0, that will be 0 if and only if dA/dx is. So as far as finding x that maximizes A is concerned, it is the same as the x that maximizes A^2. And since A^2 does not have that square root, it is easier to work with. MarkFL: You will find this a useful "shortcut" when optimizing distances as well. Friar: @ HallsofIvy what do you mean by simplicity in this case. you mean the author just wanted to study the effect. He was not inter-ested in actual calculations. Ok the author took the square on the RHS. Not the left hand side. In mathematics (equations) the left hand side=Right hand side. Let A = square of Area i.e. A = (Length X width)2 I think you would say that the author is supposing (letting) that A is equal to the square of Area. Then it looks ok. @MarkFL: please explain what you mean by optimizing. Do Optimizing is applied in those situations where you want to study the effect of certain mathematical phenomenon.in this case maximizing and minimizing.

 

[HallsofIvy]

You have, in your original post, A= 2xy, with \(\displaystyle y= \sqrt{100- x^2}\), so that \(\displaystyle A= 2x\sqrt{100- x^2}\). Now, you can differentiate that directly by writing it as \(\displaystyle A= 2(100x^2- x^4)^{1/2}\) and using the power rule as well as the chain rule: \(\displaystyle A'= 2(1/2)(100x^2- x^4)^{-1/2}(200x- 4x^3)= \frac{200x- 4x^3}{\sqrt{100x^2- x^4}}\). Setting that equal to 0, since a fraction is 0 if and only if the numerator is 0, we get must have \(\displaystyle 200x- 4x^3= 4x(50- x^2)= 0\) so that x= 0 or \(\displaystyle x= \sqrt{50}= 5\sqrt{2}\).

If, instead, we work with \(\displaystyle A^2= 4x^2(100- x^2)= 400x^2- 4x^4\), we have \(\displaystyle 2AA'= 800x- 16x^3\). Since "2A" is not 0, that will be 0 only if A'= 0 and that is where \(\displaystyle 800x- 16x^3= 16x(50- x^2)= 0\) again giving x= 0 or \(\displaystyle x= 5\sqrt{2}\).