SOLUTION: An architect is designing an atrium for a hotel. The atrium is to be rectangular with a perimeter of 688 ft of brass piping. What dimensions will maximize the area of the​ at

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: An architect is designing an atrium for a hotel. The atrium is to be rectangular with a perimeter of 688 ft of brass piping. What dimensions will maximize the area of the​ at      Log On


   

Question 1058527: An architect is designing an atrium for a hotel. The atrium is to be rectangular with a perimeter of 688 ft of brass piping. What dimensions will maximize the area of the​ atrium?
Found 2 solutions by ikleyn, josmiceli:
Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
A rectangle with the given perimeter which has maximal area is a SQUARE.

To find the side length of the square divide 688 ft by 4: side = 688%2F4 = 172 ft.


See the lesson
    - A rectangle with a given perimeter which has the maximal area is a square
in this site.


Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this textbook under the topic "Finding minimum/maximum of quadratic functions".


The other lessons under this topic are
    - HOW TO complete the square to find the minimum/maximum of a quadratic function
    - Briefly on finding the minimum/maximum of a quadratic function
    - HOW TO complete the square to find the vertex of a parabola
    - Briefly on finding the vertex of a parabola
    - A farmer planning to fence a rectangular garden to enclose the maximal area
    - A farmer planning to fence a rectangular area along the river to enclose the maximal area
    - A rancher planning to fence two adjacent rectangular corrals to enclose the maximal area



Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Formula for perimeter of rectangle:
+P+=+2W+%2B+2L+
+P+=+688+ ft
+688+=+2W+%2B+2L+
+344+=+W+%2B+L+
+L+=+344+-+W+
---------------------
Formula for area of a rectangle:
+A+=+W%2AL+
+A+=+W%2A%28+344+-+W+%29+
+A+=+-W%5E2+%2B+344W+
-------------------------
The formula for +W%5Bmax%5D+ is
+W%5Bmax%5D+=+-b%2F%282a%29+
+a+=+-1+
+b+=+344+
+W%5Bmax%5D+=+-344%2F%282%2A%28-1%29+%29+
+W%5Bmax%5D+=+172+
+L%5Bmax%5D+=+344+-+W+
+L%5Bmax%5D+=+172+
------------------------------
Plug this result back into equation
+A%5Bmax%5D+=+-172%5E2+%2B+344%2A172+
+A%5Bmax%5D+=+-29584+%2B+59168+
+A%5Bmax%5D+=+29584+
+A%5Bmax%5D+=+172%5E2%0D%0Aalso%0D%0A%7B%7B%7B+A%5Bmax%5D+=+W%2AL+
+A%5Bmax%5D+=+172%2A172+
This shows that the maximum area is a square
( which is always true )
-------------------------
Here's the plot od +A+=+-W%5E2+%2B+344W+
+graph%28+400%2C+400%2C+-100%2C+500%2C+-3500%2C+35000%2C+-x%5E2+%2B+344x+%29+