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andreev551 [17]
4 years ago
15

A local club is arranging a charter flight. The cost of the trip is $360 each for 90 passengers, with a refund of $2 per passeng

er for each seat sold in excess of 90. (Hint: x=number of passengers over 90.)
a.) Write the revenue function R(x).
b.) Find the maximum revenue and the number of passengers which will maximize the revenue. Be sure to show all necessary supporting calculus.
Mathematics
1 answer:
ludmilkaskok [199]4 years ago
3 0
A. R(x)=(360-2x)(90+x)\\R(x)=32400+360x-180x-2x^2\\R(x)=-2x^2+180x+32400

B. R(x)=-2x^2+180x+32400\\R'(x)=-4x+180=0\\180=4x\\45=x\\\\R(45)=-2(45)^2+180(45)+32400\\R(45)=-2(2025)+8100+32400\\R(45)=-4050+40500\\R(45)=36450\\\\90+45\\135

135 passengers will maximize the revenue at $36,450.
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\bold{\huge{\pink{\underline{ Solution}}}}

\bold{\underline{ Given :-}}

  • <u>A </u><u>triangle </u><u>with </u><u>sides </u><u>11m</u><u>, </u><u> </u><u>13m </u><u>and </u><u>18m</u>

\bold{\underline{ To \: Find :-}}

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\bold{\underline{ Let's \: Begin :-}}

\sf{\red{ In \:right \:angled\ : triangle, }}

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<u>We </u><u>imply</u><u> </u><u>it </u><u>in </u><u>the </u><u>given </u><u>triangle </u><u>,</u>

\sf{\red{ ( Perpendicular)² + (Base)² = (Hypotenuse)²}}

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\sf{ 121 + 169 ≠  324 }

\sf{ 290 ≠ 324  }

<u>From </u><u>Above </u><u>we </u><u>can </u><u>conclude </u><u>that</u><u>, </u>

The sum of the squares of two small sides that is perpendicular height and base is not equal to the square of longest side that is Hypotenuse

\sf{\blue{ Hence,\: Your\: answer \:is \:false }}

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