Algebra Word Problem HELPPPPP!?

A Norman window is a window 

with a semi-circle on top of a rectangular window. 

What should be the dimensions of the window 

to allow in as much light as possible, 

if there are only 12 feet of the frame material available.

We have the following perimeter

2x + 2y + pi x / 2 = 12 divide through by 2

x + y + (1/4)pi x = 6

(1 + (1/4)pi ) x + y = 6

y = 6 - (1 + .25pi)x

So....the total Area , A, to be maximized is

A = x * y + pi [(1/2)x]^2/2 substituting for y, we have

A = x * [ 6 - (1 + .25pi)x] + .25pix^2 /2 simplify

A = 6x - (1 + .25pi)x^2 + .125pix^2

A = 6x - x^2 - .25pix^2 + .125pix^2

A = 6x - x^2 - .125pix^2

A= 6x - ( 1 + .125pi)x^2

This might be most easily solved with a graph: https://www.desmos.com/calculator/brhi8ngjzf

The value of x that maximizes the area ≈ 2.154 ft

And y = [6 - (1 + .25pi)(2.154) ] ≈ 2.154 ft

So the rectangular part = 2.154 ft x 2.154 ft

Corrected Answer

3 x 3 that would be 12'

i assume that means maximize the area with a fixed perimeter.

if W = width and L = length of the straight part

A = LW + (1/2)(W/2)²π

A = LW + W²π/8

   (rectangular part plus semicircle)

perimeter = 12 = 2L + W + Wπ = 2L +W(1+π)

   (3 straight sides plus half the circumference of the circle part)

eliminate one of the variables from the area equation

12 = 2L +W(1+π)

L = 6 – W(1+π)/2

A = LW + W²π/8

A = (6 – W(1+π)/2)W + W²π/8

A = 6W – W²(1+π)/2  + W²π/8

A = 6W – W²(1/2 + π/2 + π/8)

A = 6W – W²(1/2 + 5π/8)

to get max, differentiate and set equal to zero. slope is zero at peak.

dA/dW = 6 – 2W(1/2 + 5π/8) = 0

2W(1/2 + 5π/8) = 6

W(1/2 + 5π/8) = 3

W = 3 / (1/2 + 5π/8) = 1.218 m

L = 6 – W(1+π)/2 = 6 – 1.281(1+π)/2 = 3.478

check

radius = 1.218/2 = 0.609

circular portion = 2π0.609 = 3.826

perimeter = 3.826 + 3.478 + 3.478 + 1.218 = 12.000