Urgent Help: Optimization Question

Question

A building is to be built and will take the form of a semicircular shape and will have a rectangular doorway. The building will be 6 m long and the door must be at least 1 m vertically from the roof of the building.

A) What is the largest area the doorway can have and what are its dimensions(give the area to the nearest 0.01 m^2 and the dimensions to the nearest cm)

 

Can someone please fully explain the steps? Including the Product rule and Simplification?

Answer 1

The equation of the circle is X² + Y² =9

The two critical points are the upper corners of the door, there the minimum distance must be 1M to the roof, so the X coordinate when Y=2 is => X=√5.

Now we need to make a choice, the first is to say that the height of the door is 2 and the width will be 2√5

giving an area equal to 4√5 = 8.94 m²

But in this case the door will be in the horizontal sense touching the roof, I have been working in an equation where the distance from x is 1 and from y is 1.

Answer 2

Hi Lena,

 

This question is a little difficult to explain without the use of diagrams, but I will try my best to get it through to you.

 

Draw a semi circle centered on the origin. Then, draw a rectangle inscribed within the semi circle, i.e. the corners of the rectangle are touching the semi circle.

 

Since the base of the building is 6m long, the radius of the semi circle is 3. Hence, we can describe the circle as

 

x² + y² = 3²

 

x2 + y2 = 9

 

Further, since the doorway must be at least 1m vertically from the roof of the building, 0≤y≤2. Our constraints for the optimization problem are thus

 

x² + y² = 9

 

and

 

0 ≤ y ≤ 2

 

Now, we can begin to work on our optimization. Say the bottom of the doorway is x. Since our semi circle is centered on the origin, x only covers half of the base of the doorway, so the base of the doorway is 2x. Let y represent the height of the doorway. Then,

 

A = (2x)y

 

From our constraint x² + y² = 9, we have

 

y² = 9 - x²

 

or

 

y = √(9 - x²)

 

Plugging this into our equation for the area of the doorway, we have

 

A = 2xy

 

   = 2x √(9 - x²)

 

   = 2x(9 - x²)^1/2

 

Then, we take the derivative

 

dA/dx = 2x ((1/2)(9 - x²)^1/2-1(-2x)) + 2(9 - x²)^1/2

 

          = -2x²/√(9 - x²) + 2√(9 - x²)

 

          = ( -2x² + 2√(9 - x²)√(9 - x²) )/√(9 - x²)

 

          = ( -2x² + 2(9 - x²) ) / √(9 - x²)

 

          = ( -4x² + 18 ) / √(9 - x²)

 

Setting this equal to zero amounts to solving

 

-4x² + 18 = 0

 

or

 

18 = 4x²

 

Solving this for x yields x = √(9/2). To find y, we plug x = √(9/2) into our constraint

 

x² + y² = 9

 

(√(9/2))² + y² = 9

 

or just

 

y² = 9 - 9/2

 

    = 9/2

 

So, x = y = √(9/2) which is approximately 2.12.

 

However, since the top of the doorway must be at least 1m away from the roof of the shed, this optimal value for y will not work as 2.12 is 0.88m away fromt the roof. Our only other choice for y is y = 2. If y = 2, then

 

x² + 2² = 9

 

yields

 

x² = 9 - 4

 

and so

 

x = √5

 

The area is then

 

A = 2xy

 

   = 2(√5)(2)

 

   = 4√5

 

To the nearest 0.01 m², A = 8.94 m², base = 2x = 4.47 m, and height = y = 2m.