Maxima and Minima problem (2 Viewers)

[cutemouse]

Hello guys, I'm having trouble solving this question. Could someone please help me?

A large bay window is to have the shape of a rectangle topped by a semi-circle as shown in the diagram below (attached). The rectangle is of clear glass while the semi-circle is of tinted glass that transmits only half as much light per unit area as the clear glass does.

If the total perimeter of the window is 20m, find the dimensions x and y of the window admitting the most light.

Thanks guys

 

[gurmies]

Perimeter = x + 2y + (πx)/2 = 20

2x + 4y + πx = 40

Now for the area, you have to realise that (xy)/2 = (πx²)/8 cause it's in terms of the light that they "admit".

So total Area = (xy)/2 + (πx²)/8

From the perimeter, we can say that y = 10 - (πx)/4 - x/2

A = [x(10 - (πx)/4 - x/2]/2 + (πx²)/8

= (10x - πx²/4 - x²/2)/2 + (πx²)/8

= (40x - πx² - 2x²)/8 + (πx²)/8

= 5x - x²/4

dA/dx = 5 - x/2

At stationary points, dA/dx = 0

5 - x/2 = 0

x/2 = 5

x = 10

d²A/dx² = -1/2

Since d²A/dx² < 0 when x = 10, it's a maximum.

When x = 10, y = 5 - 5π/2 which seems impossible now that I look at it, y is a negative -_-

 

[cutemouse]

It's light transmitted per unit area... so shouldn't I let the outside intensity of light be I and then divide it by the area, and then equate them using the 1:0.5 factor?

 

[lolokay]

light, L, is proportional (by constant k) to rectangle area + half of semicircle area
L/k = xy + πx²/16
and x + 2y + (πx)/2 = 20

[sub in for y, differentiate, solve for max]

hopefully gives x = 10/(1+3π/8)

 

[gurmies]

Ahh, I see =)

 

[cutemouse]

Oh crap... I'm lost...

Could you please explain that to me about the constant and L?