Problem on Maximisation and Minimisation (1 Viewer)

[GaDaMIt]

12. a piece of wire length 80cm is to be cut into two sections. one section is to be bent into a square, and the other into a rectangle 4 times as long as it is wide
a) let x be the side lengh of the square and y be the width of the rectangle ... basically had to prove sum of areas of square - rectangle is given as

A = (41/4)y^2 - 100y + 400
.. ive done that

b) Find the lengths of both sections of wire if A is to be a minimum .. dont get how to do this



EDIT:
Might as well ask another question which im only getting half the answer for....


(x+5)/(x-5) - (x-6)/(x+6) = (x+4)/(x-4) - (x-7)/(x+7)

i get down to 2x+2 = 0
x= -1.. but answers have x = -1 and 0

 

[SoulSearcher]

Differentiate A with respect to y, then find y when dA/dy = 0, then show that the y value you just figured out is a minimum by finding d²A/dy², then find the perimetre of the rectangle, then subtract that length from 80 to find the perimeter of the square.

 

[Riviet]

(x+5)/(x-5) - (x-6)/(x+6) = (x+4)/(x-4) - (x-7)/(x+7)

i get down to 2x+2 = 0
x= -1.. but answers have x = -1 and 0

You should have reached a line something like this:

x(x-4)(x+7) = x(x-5)(x+6)

DO NOT cancel the x's on both sides, by doing so means you have lost a solution of x=0. Strictly speaking though, you mathematically aren't allowed to cancel x because x COULD be zero (and in this case it is!) and you can't divide by zero.