document.write( "Question 166298: A snack bar cooks and sells hamburgers and hotdogs during football games. To stay in business, it must sell at least 10 hamburgers, but cannot cook more than 40. It must also sell at least 30 hotdogs, but cannot cook more than 70. The snack bar cannot cook more than 90 items total. The profit on a hamburger is $0.33 and on a hotdog it is $0.21. How many of each item should it sell to make the maximum profit? \n" ); document.write( "

Algebra.Com's Answer #122564 by Mathtut(3670) $\"\"$ $\"About$

You can put this solution on YOUR website!
I am having problems mathematically showing this \r
\n" ); document.write( "\n" ); document.write( "but we know that $\"10%3C=h%3C=40\"$
\n" ); document.write( "and............. $\"30%3C=d%3C=70\"$
\n" ); document.write( "and............. $\"h%2Bd%3C=90\"$ where h= # of hamburgers and d= # of hotdogs.\r
\n" ); document.write( "\n" ); document.write( "when one takes the max # of hamburgers which make more profit
\n" ); document.write( "than dogs we have 40 burgers meaning we can only have 50 dogs
\n" ); document.write( "because $\"h%2Bd%3C=90\"$ ...so .33(40)+.21(50)= $23.70 profit....now if you take the maximum # of dogs you would have 70 meaning you can only have 20 hamburgers again due to $\"h%2Bd%3C=90\"$ . so we have .33(20)+.21(70)= $21.30
\n" ); document.write( "profit. anything combo of hamburgers and hotdogs between these maximums also render profits that are between those 2 profits so my opinion is that this is the correct answer. $\"40-hamburgers-and-50-hotdogs\"$ \n" ); document.write( "

\n" );