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. What are the dimensions of the rectangle with maximum area?

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)

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.

10 down (bang1)

5 up

5down (2)

2.5 up

2.5 down (3)

1.25 up

1.25 down (4)

now add them up : )

hoped i lol