Difference between revisions of "Math 22 Optimization Problems"

Revision as of 06:24, 1 August 2020

Find Maximum Area: Find the length and width of a rectangle that has 80 meters perimeter and a maximum area.

Solution:
Let $l$ be the length of the rectangle in meter.
and $w$ be the width of the rectangle in meter.
Then, the perimeter $P=2(l+w)=80$ , so $l+w=40$ , then $l=40-w$
Area $A=l.w=(40-w)w=40w-w^{2}$
$A'=40-2w=0$ , then $w=20$ , so $l=40-w=40-20=20$
Therefore, $l=w=20$

This page were made by Tri Phan

@@ Line 1: / Line 1: @@
-==Solving Optimization Problems=
+==Solving Optimization Problems==
-'''Find Maximum Area''': Find the length and width of a rectangle that has 80 meters perimeter.
+'''Find Maximum Area''': Find the length and width of a rectangle that has 80 meters perimeter and a maximum area.
+{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
+!Solution: &nbsp;
+|-
+|Let <math>l</math> be the length of the rectangle in meter.
+|-
+|and <math>w</math> be the width of the rectangle in meter.
+|-
+|Then, the perimeter <math>P=2(l+w)=80</math>, so <math>l+w=40</math>, then <math>l=40-w</math>
+|-
+|Area <math>A=l.w=(40-w)w=40w-w^2</math>
+|-
+|<math>A'=40-2w=0</math>, then <math>w=20</math>, so <math>l=40-w=40-20=20</math>
+|-
+|Therefore, <math>l=w=20</math>
+|}
-'''Solution''':