Answer:
The length and width that maximize the area are:
W = 2*√8
L = 2*√8
Step-by-step explanation:
We want to find the largest area of a rectangle inscribed in a semicircle of radius 4.
Remember that the area of a rectangle of length L and width W, is:
A = L*W
You can see the image below to see how i will define the length and the width:
L = 2*x'
W = 2*y'
Where we have the relation:
4 = √(x'^2 + y'^2)
16 = x'^2 + y'^2
Now we can isolate one of the variables, for example, x'
16 - y'^2 = x^'2
√(16 - y'^2) = x'
Then we can write:
W = 2*y'
L = 2*√(16 - y'^2)
Then the area equation is:
A = 2*y'*2*√(16 - y'^2)
A = 4*y'*√(16 - y'^2)
If A > 1, like in our case, maximizing A is the same as maximizing A^2
Then if que square both sides:
A^2 = (4*y'*√(16 - y'^2))^2
= 16*(y'^2)*(16 - y'^2)
= 16*(y'^2)*16 - 16*y'^4
= 256*(y'^2) - 16*y'^4
Now we can define:
u = y'^2
then the equation that we want to maximize is:
f(u) = 256*u - 16*u^2
to find the maximum, we need to evaluate in the zero of the derivative:
f'(u) = 256 - 2*16*u = 0
u = -256/(-2*16) = 8
Then we have:
u = y'^2 = 8
solving for y'
y' = √8
And we know that:
x' = √(16 - y'^2) = √(16 - (√8)^2) = √8
And the dimensions was:
W = 2*y' = 2*√8
L = 2*y' = 2*√8
These are the dimensions that maximize the area.
