Determine the dimensions of the rectangle of largest area that can be inscribed in a semicircle of radius 4

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.

i think it would be 42cm

they used 313 gallons of paint