Question

Find the area of the largest rectangle that
can be inscribed in a semi-circle of radius 3.

Answer 1

using calculus?

Answer 2

yepp

Answer 3

ok the equation of a circle that is centered at the origin with radius 3 is
x^2 + y^2 = 3^2

Answer 4

if you put a rectangle in there, it will have coordinates ( -x, sqrt (9-x^2)) and
(x, sqrt (9-x^2)), (x,0) and (-x,0)

Answer 5

the area of the rectange is
A = 2x * y , where y = sqrt (9 -x^2)

Answer 6

A' = 2 sqrt (9-x^2) + 2x (1/2) ( 9-x^2)^(-1/2)(-2x)

Answer 7

ohhh. i thought the area was 2x2y which made it 4xy

Answer 8

A = 2sqrt(9-x^2) - x^2/ sqrt (9-x^2)

Answer 9

common denominator that gives us
A = [ 2(9-x^2) - x^2] / ... , and we dont care about denominator,
so 18-3x^2 = 0

Answer 10

so x = sqrt 6 is your max for x, so your base will have 2sqrt 6 , and y = sqrt 3

Answer 11

so i get 12

Answer 12

(l/2)^2+(w/2)^2=3^2=9 (1)
and u know that A=lw for rectangle so solve equation (1) in terms of length and substitute into this area equation
l=sqrt(9-(w/2^2)) so A=lw=(sqrt(9-(w/2^2))(w) then take derivative of this and set it equal to zero and solve for W which will give you maximum width
dA/dw=(1/2)(9-(w/2^2))*(w^2)+(1/2)(9-(w/2^2)(w)=0 solve for w

Answer 13

how? the answer choices are 9, 6, 5, 8, and 7 sq. units

Answer 14

and once u solve for that then use equation (1) to solve for corresponding length...

Answer 15

i found an error

Answer 16

A = [ 2(9-x^2) -2 x^2] / sqrt 9-x^2

Answer 17

so max x = 3/sqrt 2

Answer 18

= 3 sqrt 2 / 2

Answer 19

now we want 2x , so the base is 3 sqrt 2

Answer 20

uhh

Answer 21

josee are you boy or girl

Answer 22

i got lost looking for w, and i dont understand your last step fermats

Answer 23

um boy. why?

Answer 24

here is the solution in a pdf http://science.kennesaw.edu/~plaval/math1190/maxmin.pdf

Answer 25

scroll down

Answer 26

the answer is 9

Answer 27

0 Find the area of the largest rectangle that can be inscribed in a
semi circle of radius r.
For simplicity, we consider the upper half of the circle of radius r, centered at
the origin. Its equation is x
2 + y
2 = r
2
with y 0. Thus y =
p
r
2 x
2
. We
label (x; y) the upper right corner of the rectangle. Thus, the rectangle has height
y and width 2x. Let A denote the area of this rectangle. Then, A = 2xy. We
want to maximize A. First, we express A as a function of one other quantity.

Answer 28

for sum reason i don't feel like the answer is 9 if it it is 9 then what did u get fermet for l and w ?

Answer 29

She said x=3/sqrt(2), so w=6/sqrt(2) and h=3/sqrt(2).