Expert: Socrates - 6/21/2007

Question
a window has the shape of a rectangle surmounted by an isosceles triangle.  If the perimeter of the window is 12 feet, what values x y and è will maximize the total area? (è is the angle at the left, so i suppose is also the same angle for the right)

Answer
I will assume that the length and width of the rectangle are x and y.

The area of the triangle is (1/4)x^2 tan(è)

The area of the rectangle is xy

The area of the window is then

Area = (1/4)x^2 tan(è) + xy

The perimeter is x + 2y + (x)sec(è)

We know the perimeter is 12 , so

12 = x + 2y + (x)sec(è)

Solving for y , this gives

y = 6 - (x/2)(1+sec(è))

Substitute this expression for y into the expression for Area and get

Area = (1/4)x^2 tan(è) + 6x - (x^2/2)(1+sec(è))

Simplifying ,

Area = [(1/4)tan(è)-(1/2)(1+sec(è))] x^2 + 6x

The maximum must occur when the partial derivatives with respect to x and è of this expression are both 0

The derivative with respect to x is

(2)[(1/4)tan(è)-(1/2)(1+sec(è))] x + 6

The derivative with respect to è is

[(1/4)(sec(è))^2 -(1/2)(sec(è))(tan(è))] x^2


Set the last expression equal to 0 ,

[(1/4)(sec(è))^2 -(1/2)(sec(è))(tan(è))] x^2 = 0

Since x can not be zero, we can divide both sides by x^2 and get

[(1/4)(sec(è))^2 -(1/2)(sec(è))(tan(è))] = 0


multiply through by 4 to clear the fractions

[(sec(è))^2 - (2)(sec(è))(tan(è))] = 0

multiply through by (cos(è))^2 and get

[ 1 - (2)(sin(è))] = 0

then

sin(è) = 1/2


From this we find that è = pi/6 or 30 degrees


We also have the derivative with respect to x equal to zero,


(2)[(1/4)tan(è)-(1/2)(1+sec(è))] x + 6 = 0

Substitute è = pi/6 into this equation and find

(2)[(1/4)(3^(1/2)/3)-(1/2)(1+(2)(3^(1/2)/3)] x + 6 = 0

Multiply through by 4 to simplify


(2)[(3^(1/2)/3)- (2)(1+(2)(3^(1/2)/3)] x + 24 = 0

Divide both sides by 2

[(3^(1/2)/3)- (2)(1+(2)(3^(1/2)/3)] x + 12 = 0

Multiply both sides by 3

[3^(1/2)- (2)(3+(2)(3^(1/2))] x + 36 = 0

Simplify this to get

[(-3)(3^(1/2)) - 6 ] x + 36 = 0

Divide through by -3

[ 3^(1/2) + 2 ] x - 12 = 0

now solve for x

x = 12 /[ 3^(1/2) + 2 ]


To find the maximum area , these values


è = pi/6 and x = 12 /[ 3^(1/2) + 2 ]

must be substituted into the expression for area


Area = [(1/4)tan(è)-(1/2)(1+sec(è))] x^2 + 6x


After making these substitutions and simplifying , we find


Area = 36/( 2 + 3^(1/2))