bmwhype2
so...
if we want to
maximize the area of a figure that has two adjacent sdies of 10, we go with the square rather than the rhombus?
Yes... My view on it under... (Out of GMAT)
Let:
> h be the height of a rombhus
> w be the width of a rombhus
o (w/2)^2 + (h/2)^2 = 10^2
<=> w^2 + h^2 = (20)^2
=> 2*w + d(h^2)/dw = 0..... we derivate by dw
<=> 2*w + 2*h*dh/dw = 0
<=> w = -h * dh/dw
<=> w/-h = dh/dw
Then, the area is such:
o Area = (h)*(w)/2
The maximum of area is obtained for an extremum, implying that if we derivate by dw or dh, we obtain 0.
dArea/dw = 0
<=> w*dh/dw + h*dw/dw = 0
<=> w*w/-h + h*dw/dw = 0
<=> -w^2 + h^2 = 0
<=> w^2 = h^2
<=> w = h
The square is the optimized rombhus to contain the most