SOLUTION: (BEGINNING OF QUESTION)
Mike and Bill's ages if added together equal 9. The product of one of their ages multiplied by the square of the other's age is a maximum. What are the pos
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Question 835084: (BEGINNING OF QUESTION)
Mike and Bill's ages if added together equal 9. The product of one of their ages multiplied by the square of the other's age is a maximum. What are the possible combinations of their ages?
Take your time to logically lay out your thoughts as to the process and statement of equations you would use to solve the problem. You will need the use of quadratic equations and their solutions, you will also need to take a derivative of an equation(s)and set its result to zero to determine a maximum value.
(END OF QUESTION)
(MY ATTEMPT)
I tried different combinations of their ages and got 3(6^2) as the maximum amount you can get from ages that add up to 9. The only problem is "The professor wants us to include the quadratic formula and some sort of derivation". I don't even understand where to apply the formula and the derivation.
Thanks for the help.
Found 2 solutions by josgarithmetic, richwmiller:
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
m and b, their ages mike and bill.
First sentence, in symbols, .
Maximum? m*b^2 is a maximum? This could be a function.
; we can substitute for either variable using the age sum equation. Try, . This gives ,
, which now seems to be a calculus problem, since you want the maximum.
(Take derivative, find where derivative equals zero)
(The derivative will be a quadratic function)
-
Using the graphing capability of Google, a maximum appears to occur at b=6, therefore m=3.
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
m+b=9,
m=9-b
m*b^2=y
(9-b)*b^2=y
9 b^2-b^3 = y
derivative
18b-3b^2=0
18b=3b^2
18=3b
6=b
max{(9-b) b^2 = y} = 108 at b = 6
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