Math 22 Optimization Problems - Revision history

Tphan046 at 16:47, 1 August 2020

2020-08-01T16:47:42Z

2020-08-01T16:47:33Z

Solving Optimization Problems

2020-08-01T16:46:32Z

2020-08-01T16:45:52Z

Solving Optimization Problems

2020-08-01T16:14:49Z

2020-08-01T14:24:56Z

=Solving Optimization Problems

2020-08-01T14:11:48Z

2020-07-31T15:31:03Z

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← Older revision		Revision as of 16:47, 1 August 2020
Line 1:		Line 1:
−	==Solving Optimization Problems==	+	==Solving Optimization Sample Problems==
	'''1) Maximum Area''': Find the length and width of a rectangle that has 80 meters perimeter and a maximum area.		'''1) Maximum Area''': Find the length and width of a rectangle that has 80 meters perimeter and a maximum area.

@@ Line 53: / Line 53: @@
 |<math>F'=2-\frac{18000}{a^2}=0</math>, then <math>a^2=9000</math>, so <math>a=\sqrt{9000}=\pm 30=30</math> since <math>a</math> is positive. Then, <math>b=\frac{180000}{30}=6000</math>
 |-
-|Therefore, the dimensions of the fence is
+|Therefore, the dimensions of the fence is <math>a=30</math> meters and <math>b=6000</math> meters.
 |}

@@ Line 1: / Line 1: @@
 ==Solving Optimization Problems==
-'''Find Maximum Area''': Find the length and width of a rectangle that has 80 meters perimeter and a maximum area.
+'''1) 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;"
@@ Line 18: / Line 18: @@
 |}

← Older revision		Revision as of 14:24, 1 August 2020
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:
		+	\|-
		+	\|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''':~~

@@ Line 55: / Line 55: @@
 |Therefore, the dimensions of the fence is <math>a=30</math> meters and <math>b=6000</math> meters.
 |}
 [[Math_22| '''Return to Topics Page''']]
 '''This page were made by [[Contributors|Tri Phan]]'''