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September 16th, 2007, 04:56 PM  #1 
Newbie Joined: Sep 2007 Posts: 3 Thanks: 0  Can You Figure This Out?
Find 2 numbers whose sum is 24 and whose product is a maximum.

September 16th, 2007, 05:27 PM  #2 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0 
Let us say that there are two numbers a and b. a+b=24 Now, since we want to make the product ab the maximum, 24<a<0 and 24<b<0, since we want the product maximum, we can also therefore "assume" that a+b=24, and let 0<a<24 and 0<b<24. Using ArithmeticGeometric equation, a+b>=2sqrt(ab) 24>=2sqrt(ab) 12>=sqrt(ab) Since ab=144 makes the equation at top true, using algebra a(24a)=144 a^224a+144=0 (a12)^2=0 Therefore, a=12 and b=12, which the two numbers are the final answer. 
September 16th, 2007, 05:38 PM  #3 
Newbie Joined: Sep 2007 Posts: 3 Thanks: 0 
Oh now i get it. Thanks. I have another one that I am confused about. Here it is... A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. He does not need a fence along the river. What are the dimensions of the field of largest area that he can fence? Also, Find a function that models the area of the field in terms of one of its sides. 
September 16th, 2007, 05:45 PM  #4 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0 
Let the length of the rectangular field be a, and let the width of the rectangular field be b. 2400 ft=2a+2b 1200 ft=a+b We want the maximum value of ab. ab=a(1200fta)=A (which this is an equation for the "area" of the rectangular field) A=a*1200ft  a^2 Take the derivative of the area equation, and set it to zero, then use second derivative test to make sure that this will be the maximum value for the area of the rectangle A'=1200ft  2a=0 2a=1200ft a=600ft A"=2 (this tells us that the area function will always be concave down, thus when derivative equals zero, then that part will be the absolute maximum) Therefore, 600 ft by 600 ft of fencing is required to make the area largest, and is correct according to the perimeter of the geometric dimension. 
September 16th, 2007, 07:04 PM  #5 
Newbie Joined: Sep 2007 Posts: 9 Thanks: 0 
Thank you Johnny for just being you.

September 16th, 2007, 07:39 PM  #6  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
Maximize: ab Maximize: a(2400  2a) Maximize: 2400a  2aa Zero: 2400  4a Solve: 2400 = 4a a = 600 Substitution gives b = 1200.  
September 17th, 2007, 07:10 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,937 Thanks: 2210 
For the first problem, using a and b for the two numbers, ab = ((a + b)²  (a  b)²)/4 = ((24)²  (a  b)²)/4 is maximized when a = b, i.e., when a = b = 12 (since a + b = 24). The second problem can similarly be solved without use of calculus. 

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