Maximizing the Sum of Two Numbers: An Algebraic Solution

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To find two numbers that sum to 20 and maximize their product, the first number is represented as X and the second as 20-X. The product function is X(20-X), which simplifies to -X^2 + 20X, forming a parabola. The maximum product occurs at the vertex of the parabola, calculated using the axis of symmetry formula, yielding X = 10. Thus, the two numbers that maximize the product are both 10.
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Homework Statement


find two numbers whose sum is 20 and whose product is a maximum.


Homework Equations


the first number is X
the second number is 20-x



3. The solution
X(20-X)=0
-X^2+ 20x=0
x=-b/2a = -20/2(-1) = 10
20 - x =20 -10 = 10

the numbers are 10 and 10


i just don't get why / how you know to put x and 20 - x and why you would use the axis of symmetry to find the numbers

and sry mods i posted originally in the wrong thread.
 
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From the problem statement you have 2 numbers which sum to 20, that is x and 20-x.

It should be obvious that ( x )+ (20 -x) = 20 so you have represented the 2 numbers in general. Now you need to find when the product x(20-x) is a maximum.

Now if you were in a calculus class you would take the derivitive and set it to zero. Since you are not doing this I will have to assume that you are not in calculus. You have the problem of finding the maximum of the parabola, using properties of a parabola. The maximum will lie on the axis of symetry of the parabola, so all you need do is find the point on the parabola which lies on the symetry axis.
 
thanks
not in calc but next year trig then pre and then calc
 
X(20-X)=0 is not true. You have the function 20X- X2 which is a parabola with maximum value at its vertex. You can find the (X,Y) coordinates of the vertex by completing the square.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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