Answer:
a. A = l(2 - l)
b. l = 1 foot and w = 1 foot.
Step-by-step explanation:
With a constant perimeter of 4 feet, a rectangle has variable length l and variable width w.
So, 2(l + w) = 4
⇒ l + w = 2 .........(1)
⇒ w = 2 - l ........(2)
a. Now, area of the rectangle will be A = lw = l(2 - l) ......... (3) {From equation (2)}
b. For A to be maximum, the condition is [tex]\frac{dA}{dl} =0[/tex].
Now, from equation (3), differentiating with respect to l, we get
[tex]\frac{dA}{dl} =2-2l = 0[/tex].
⇒ l = 1 feet.
Hence, w = 1 feet.
So, when the length and width of the rectangle are the same and equal to 1 foot, then only the area will be maximum.
That means, when the rectangle becomes a square, then the area will be maximum.
