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Difference between revisions of "2003 AMC 10B Problems/Problem 4"

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==Problem 4==
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{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #3]] and [[2003 AMC 10B Problems|2003 AMC 10B #4]]}}
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==Problem==
  
 
Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the figure. She plants one flower per square foot in each region. Asters cost <math> \ </math><math>1</math> each, begonias <math> \ </math><math>1.50</math> each, cannas <math> \ </math><math>2</math> each, dahlias <math> \ </math><math>2.50</math> each, and Easter lilies <math> \ </math><math>3</math> each. What is the least possible cost, in dollars, for her garden?
 
Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the figure. She plants one flower per square foot in each region. Asters cost <math> \ </math><math>1</math> each, begonias <math> \ </math><math>1.50</math> each, cannas <math> \ </math><math>2</math> each, dahlias <math> \ </math><math>2.50</math> each, and Easter lilies <math> \ </math><math>3</math> each. What is the least possible cost, in dollars, for her garden?
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==Solution==
 
==Solution==
  
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The areas of the regions from greatest to least are <math>21,20,15,6</math> and <math>4</math>.
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The areas of the five regions from greatest to least are <math>21,20,15,6</math> and <math>4</math>.
  
 
If we want to minimize the cost, we want to maximize the area of the cheapest flower and minimize the area of the most expensive flower. Doing this, the cost is <math>1\cdot21+1.50\cdot20+2\cdot15+2.50\cdot6+3\cdot4</math>, which simplifies to <math> \ </math><math>108</math>.
 
If we want to minimize the cost, we want to maximize the area of the cheapest flower and minimize the area of the most expensive flower. Doing this, the cost is <math>1\cdot21+1.50\cdot20+2\cdot15+2.50\cdot6+3\cdot4</math>, which simplifies to <math> \ </math><math>108</math>.
 
Therefore the answer is <math>\boxed{\textbf{(A) } 108}</math>.
 
Therefore the answer is <math>\boxed{\textbf{(A) } 108}</math>.
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==See Also==
 
==See Also==
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{{AMC12 box|year=2003|ab=B|num-b=2|num-a=4}}
 
{{AMC10 box|year=2003|ab=B|num-b=3|num-a=5}}
 
{{AMC10 box|year=2003|ab=B|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 23:11, 4 January 2014

The following problem is from both the 2003 AMC 12B #3 and 2003 AMC 10B #4, so both problems redirect to this page.

Problem

Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the figure. She plants one flower per square foot in each region. Asters cost $$$1$ each, begonias $$$1.50$ each, cannas $$$2$ each, dahlias $$$2.50$ each, and Easter lilies $$$3$ each. What is the least possible cost, in dollars, for her garden?

[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((6,0)--(0,0)--(0,1)--(6,1)); draw((0,1)--(0,6)--(4,6)--(4,1)); draw((4,6)--(11,6)--(11,3)--(4,3)); draw((11,3)--(11,0)--(6,0)--(6,3)); label("1",(0,0.5),W); label("5",(0,3.5),W); label("3",(11,1.5),E); label("3",(11,4.5),E); label("4",(2,6),N); label("7",(7.5,6),N); label("6",(3,0),S); label("5",(8.5,0),S);[/asy]

$\textbf{(A) } 108 \qquad\textbf{(B) } 115 \qquad\textbf{(C) } 132 \qquad\textbf{(D) } 144 \qquad\textbf{(E) } 156$

Solution

The areas of the five regions from greatest to least are $21,20,15,6$ and $4$.

If we want to minimize the cost, we want to maximize the area of the cheapest flower and minimize the area of the most expensive flower. Doing this, the cost is $1\cdot21+1.50\cdot20+2\cdot15+2.50\cdot6+3\cdot4$, which simplifies to $$$108$. Therefore the answer is $\boxed{\textbf{(A) } 108}$.

See Also

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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