Find the dimensions of the rectangle of maximum area that can be formed from a 50-in piece of wire.
What are the dimensions and maximum area?
- PhilipLv 62 months ago
Let (length,width) of rectangle = (x,y).
Perimeter of rectangle = 2(x+y) = 50 and y = 25-x.
Area of rectangle = f(x) = x(25-x) = 25x - x^2.
f'(x) = 25 - 2x.
f''(x) = -2.
An extremum occurs wherever f'(x) = 0.
f'(x) = 0 for x = (25/2).
f''(x) = -2, < 0, ---> extremum is a maximum.
f(25/2) = (25/2)[2(25/2) -1(25/2)] = (25/2)^2 = (625/4) = 156.25.
The rectangle is actually a square of side length (25/2).
- ted sLv 72 months ago
it is well known that the figure is a square
- rotchmLv 72 months ago
Let x & y be the width & length resp.
You Are given that 2x+2y = 50, right?
And the area is A = xy, right?
From the first eqs, solve for y. So you have y = y(x). Plug this y(x) into the second eqs. You now have the area A(x) solely as a fct of x. What is this function A(x) ?
Answer that & we will proceed.