1 / 13

The Hammer Window

The Hammer Window. What is the Question?. A Hammer Window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 30 feet, find the dimensions of the window so that the greatest possible amount of light is admitted. r. l. l. w. What Does That Mean?.

jadzia
Download Presentation

The Hammer Window

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. The Hammer Window

  2. What is the Question? A Hammer Window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 30 feet, find the dimensions of the window so that the greatest possible amount of light is admitted.

  3. r l l w What Does That Mean? We have a window that looks something like this: We must maximize the area of the window, while staying within the perimeter requirement.

  4. How Do We Do That? We took the perimeter formula for both the rectangle and the circle to come up with this unique “Hammer Window Perimeter” Formula: P =w + 2l + pr w = width of Hammer window domain of w: (0,30) l = length of Hammer window domain of l: (0,30) r= radius of semicircle on top w = 2ror r = ½ w p = 3.142 P = 30’

  5. How Do We Do That? We solved for l and substituted it into our special “Hammer Window Area” formula: A =lw + ½pr2 So now, we solve it!

  6. P =w + 2l + pr w = 2r 30 =2r + 2l + 3.142r 30 - 5.142r = 2l (30 - 5.142r)½ = l 15 - 2.571r = l A =lw + ½pr2 A(r) = (15-2.571r)2r + ½(3.142) r2 A(r) = 30r - 5.142r2 + 1.571r2 A(r) = 30r - 3.571r2 A’(r) = 30 - 7.142r Solving These Suckers!

  7. A’(r) = 30 - 7.142r 0 = 30 - 7.142r 7.142r = 30 r = 4.2’ w = 2r w = 8.4’ Classify r = 4.2 as a maximum: Look at the graph of A(r) = 30r - 3.571r2 and see that it achieves its maximum value at 4.2 Maximize Area

  8. (4.2, 63) A(r) = 30r - 3.571r2 The Graph

  9. Substitute r into “Hammer Window Perimeter” formula: P =w + 2l + pr P = 2r + 2l + pr 30 = 2(4.2) +2l + 3.142(4.2) 30 = 8.4 +2l + 13.19 30 =2l + 21.59 8.41 =2l 4.2 =l So What is the Length? And the pieces come together...

  10. What the heck does all that junk mean??? Wellllll…….. It means that the Hammer Window is 8.4’ wide. The length of the rectangle section is 4.2’, and the length of the semicircle on top is equal to the radius, which is 4.2’. Overall, the length is 8.4’. In English...

  11. r = 4.2’ Overall l = 8.4’ l = 4.2’ w = 8.4’ The Hammer Window

  12. Dr. Hammer is the coolest teacher in the WORLD!

  13. MERRY CHRISTMAS!!! andahappyHalloween! the end

More Related