Question

I'm still having trouble understanding this optimization question: "Find two positive real numbers whose sum is 100 and whose product is a maximum." Help?

Answer 1

you know you need to find 2 values such that
a + b = 100
and a * b gives the max value.
You know how to find the max point of a parabola (quadratic equation), so by substituting b with b = 100 = a you can rewrite
a * b as a * (100 - a).

so now you have f(a) = a(100-a), meaning you have a root at a = 0 and a = 100
0(100-0) = 0
100(100-100) = 0.

For a parabola the vertex is the max (or min) value of the function, so if you have a function with the root at a = {0,100}, then you can find your max value at a = ?

Makes sense ?

Answer 2

x+y = 100
and f(x)=xy
but y = 100-x
so
maximize f(x) = x(100-x) = -x^2+100x
take derivative
-2x+100
set equal to 0
-2x+100 = 0
x = 50

Answer 3

we know this is a max because we have degree 2 and negative leading coefficient.

Answer 4

all min/max problems can be written in the form
maximize f(x) subject to some constraint.
your constraint was given as x+y=100 and you are trying to maximize a product function. But since it has two variables xy we need to write y in terms of x, so we use the constraint. This is like having two equations with two variables and using substitution to solve

Answer 5

I don't think this is a calculus question, so you cannot take the derivative.

Answer 6

why do you say that?

Answer 7

well if not, instead of finding the derivative of -x^2+100, use the vertex -b/2a to get the same point, its then the same thing

Answer 8

no calc needed and the set up is the exact same thing

Answer 9

It is a calc problem. It's optimization

Answer 10

x+y = 100
and f(x)=xy
but y = 100-x
so
maximize f(x) = x(100-x) = -x^2+100x
find vertex
-b/2a = -100/2(-1)
x = 50

Answer 11

I just assumed since the question is not marked as a calculus question you should not use the derivative to find the max value.

Answer 12

either way, same thing...

Answer 13

ok - sorry

Answer 14

vertex vs derivative are the same at this point

Answer 15

yes - the problem here is to 'see' how to get the quadratic equation from the given information

Answer 16

thank you very much!

Answer 17

you still need to find y, but use your constraint and this is pretty trivial.

Answer 18

i've got it... as in I understand now. Thank you again

Answer 19

good deal, np

Answer 20

when doing other optimization problems, think of this example. It is a great basic concept, and the harder problems are the same, just a little harder to figure out what the constraint is, and what the f(x) is we want to max/min.