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Difference between revisions of "1989 AHSME Problems/Problem 6"
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== Problem ==
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If <math>a,b>0</math> and the triangle in the first quadrant bounded by the co-ordinate axes and the graph of <math>ax+by=6</math> has area 6, then <math>ab=</math>
 
If <math>a,b>0</math> and the triangle in the first quadrant bounded by the co-ordinate axes and the graph of <math>ax+by=6</math> has area 6, then <math>ab=</math>
  
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==Solution==
 
==Solution==
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Setting <math>y=0</math> we have that the <math>x-</math>intercept of the line is <math>x=6/a</math>. Similarly setting <math>x=0</math> we find the <math>y-</math>intercept to be <math>y=6/b</math>. Then <math>6=(1/2)(6/a)(6/b)</math> so that <math>ab=3</math>. Hence the answer is <math>\mathrm{(A)}</math>.
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Setting <math>y=0</math> we have that the <math>x-</math>intercept of the line is <math>x=6/a</math>. Similarly setting <math>x=0</math> we find the <math>y-</math>intercept to be <math>y=6/b</math>. Then <math>6=(1/2)(6/a)(6/b)</math> so that <math>ab=3</math>. Hence the answer is <math>\fbox{A}</math>.
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== See also ==
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{{AHSME box|year=1989|num-b=5|num-a=7}} 
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[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 06:44, 22 October 2014

Problem

If $a,b>0$ and the triangle in the first quadrant bounded by the co-ordinate axes and the graph of $ax+by=6$ has area 6, then $ab=$

$\mathrm{(A) \ 3 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 12 } \qquad \mathrm{(D) \ 108 } \qquad \mathrm{(E) \ 432 }$

Solution

Setting $y=0$ we have that the $x-$intercept of the line is $x=6/a$. Similarly setting $x=0$ we find the $y-$intercept to be $y=6/b$. Then $6=(1/2)(6/a)(6/b)$ so that $ab=3$. Hence the answer is $\fbox{A}$.

See also

1989 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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