document.write( "Question 1056798: A manufacturer of telephones determines that the profit from producing and selling x telephones is P(x) = -.01x 2 + 6x – 500 dollars. a. How many telephones should be produced to maximize the profit? b. What is the maximum profit? (This is an application of derivatives) \n" ); document.write( "

Algebra.Com's Answer #671894 by josmiceli(19441) $\"\"$ $\"About$

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There is a formula for finding the peak when
\n" ); document.write( "the equation looks like
\n" ); document.write( " $\"+f%28x%29+=+ax%5E2+%2B+b%2Ax+%2B+c+\"$ , it is:
\n" ); document.write( " $\"+x%5Bmax%5D+=+-b%2F%282a%29+\"$
\n" ); document.write( " $\"+x%5Bmax%5D+=+-6%2F%282%2A%28-.01%29%29+\"$
\n" ); document.write( " $\"+x%5Bmax%5D+=+6%2F.02+\"$
\n" ); document.write( " $\"+x%5Bmax%5D+=+300+\"$
\n" ); document.write( "----------------------------
\n" ); document.write( "You can also find the derivative of $\"+P%28x%29+\"$
\n" ); document.write( "and set it to zero to find the peak.
\n" ); document.write( "The derivative of $\"+P%28x%29+\"$ is:
\n" ); document.write( " $\"+-.02x+%2B+6+=+0+\"$
\n" ); document.write( " $\"+-.02x+=+-6+\"$
\n" ); document.write( " $\"+x+=+6%2F.02+\"$
\n" ); document.write( " $\"+x+=+300+\"$
\n" ); document.write( "---------------------
\n" ); document.write( "Selling 300 telephones maximizes profit
\n" ); document.write( " \n" ); document.write( "

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