I need to find two positive real numbers whose product is a maximum and where the sum of the first and three times the second is 60. And apparently, there're 5 steps to doing this, and I know none of them:
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step 1 1.x + 3. y = 60 step 2 y = (60-x)/3 step 3 p(x) = x((60-x)/3) step 4 p(x) = 20.x -(x^2)/3 step 5 max(p(x)) = 300 occurs when x = 30 and y = 10
Sweet, understand what the question is saying.... The Product $\displaystyle P$ is given by $\displaystyle xy=P$. The sum is given by ... what?
Well, it could be by a lot of things... 60*1, 30*2, 6*10...
Originally Posted by tutor.mathitutor step 1 1.x + 3. y = 60 step 2 y = (60-x)/3 step 3 p(x) = x((60-x)/3) step 4 p(x) = 20.x -(x^2)/3 step 5 max(p(x)) = 300 occurs when x = 30 and y = 10 You are amazing! Although step 4 didn't work right. I'll be damned if I know how you did that.
Originally Posted by BeSweeet Well, it could be by a lot of things... No...it can't. It given by "the sum of the first and three times the second" $\displaystyle \overbrace{x}^{\text{First}}\underbrace{+}_{\text{ sum}}\overbrace{3y}^{\text{3 times the second}}$ $\displaystyle x+3y=60$
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