Maximum Area...2

[nycmathguy]

Section 2.1
Question 74

Seeking a set up for part a & b.

 

[MathLover1]

you have a rectangle with sides x and y
plus semi-circle with diameter x
total area A=xy+(1/2)(x/2)^2*pi

 

[nycmathguy]

you have a rectangle with sides x and y
plus semi-circle with diameter x
total area A=xy+(1/2)(x/2)^2*pi

I'll take it from here and show my work.

 

[nycmathguy]

you have a rectangle with sides x and y
plus semi-circle with diameter x
total area A=xy+(1/2)(x/2)^2*pi

You said the total area is given by
A=xy+(1/2)(x/2)^2*pi. To express the total area as a function of x, I must somehow change y to an expression involving x.

Yes?

 

[MathLover1]

yes

 

[nycmathguy]

yes

I will work on this one tomorrow as well. On the M train now.

 

[nycmathguy]

yes

Part (a)

A=xy+(1/2)(x/2)^2*pi

P = 2x + 2y

16 = 2x + 2y

16 - 2x = 2y

(16 - 2x)/2 = y

8 - x = y

A(x) = 8x - x^2 + (1/2)(x/2)^2*pi

Correct?

I can't figure out part (b).

 

[MathLover1]

P = 2x + 2y => that is perimeter of the square
the perimeter of the window include: x, 2y, and perimeter of the semi-circle
=>P=x+2y+2x*π/2
a.
A=xy+(1/2)(x/2)^2*π

given that the perimeter of window is 16
16=x+2y+2x*π/2
16=x+2y+x*π...........solve for y
16-x-x*π=2y
y=8-(1/2)(1+π)x

substitute in area, to express it in terms of x

A=x(8-(1/2)(1+pi)x)+(1/2)(x^2/4)*π
A=-(3π* x^2)/8 - x^2/2 + 8x
A=1/8 (-4 - 3π) x^2+ 8 x

b.

max{1/8 (-4 - 3 π) x^2 + 8 x} = 128/(4 + 3π) at x = 32/(4 + 3 π) =>2.383652

y=8-(1/2)(1+π)(32/(4 + 3 π) )
y=(8 (2 + π))/(4 + 3 π) =>3.063942

dimensions:
x=2.383652
y=3.063942