Answer:
L = 2*√2
w = √2
Step-by-step explanation:
Given:
A rectangle is constructed with its base on the diameter of a semicircle with radius 2 and with its two other vertices on the semicircle.
Find:
What are the dimensions of the rectangle with maximum area?
Solution:
- Let the length and width of the rectangle be L and w respectively.
- We know that Length L lie on the diameter base. So , L < 4 and the width w is less than 2 . w < 2.
- Using the Pythagorean Theorem, we relate the L with w using the radius r = 2 of the semicircle.
r^2 = (L/2)^2 + (w)^2
sqrt (4 - w^2 ) = L / 2
L = 2*sqrt (4 - w^2 ) L < 4 , w < 2
- The relation derived above is the constraint equation and the function is Area A which is function of both L and w as follows:
A ( L , w ) = L*w
- We substitute the constraint into our function A:
A ( w ) = 2*w*sqrt (4 - w^2 )
- Now we will find the critical points for width w for which A'(w) = 0
A'(w) = 2*sqrt (4 - w^2 ) - 2*w^2 / sqrt (4 - w^2 )
0 = [2*sqrt (4 - w^2 )*sqrt (4 - w^2 ) - 2*w^2] / sqrt (4 - w^2 )
0 = 2*(4 - w^2 ) - 2*w^2
0 = -4*w^2 + 8
8/4 = w^2
w = + sqrt ( 2 ) ..... 0 < w < 2
- From constraint equation we have:
L = 2*sqrt (4 - 2 )
L = 2*sqrt(2)
