Maximizing Product: Find Two Numbers with Difference 30 for Math Help

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To find two numbers with a difference of 30 and a minimum product, the equations x - y = 30 and P = xy are established. By substituting y with x - 30, the product becomes P = x^2 - 30x. Differentiating this gives dP/dx = 2x - 30, leading to x = 15 and y = -15. A related problem involves finding two non-negative numbers that sum to 60 for maximum product, where the correct equation should be x + y = 60 instead of x - y. This highlights the importance of correctly setting up the equations for optimization problems in mathematics.
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Some math help please (Updated)

Find two numbers whose difference is 30 and whose product is a minimum. :confused:
 
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Let's say the numbers are x and y.

Then

x - y = 30

and we want the minimum value of P = xy

There are lots of ways you could do this (e.g. Lagrange multipliers,...) but the simplest way is substitution:

y = x - 30

so

P = x(x - 30) = x<sup>2</sup> - 30x

To find the minimum value of P, differentiate:

dP/dx = 2x - 30 = 0

Therefore x = 15 and y = -15
 
Thank you very much James.
 
ok ran into some trouble on a similar problem.

Find two non-negative numbers whose sum is 60 and whose product is a max.

I did the following

x-y = 60
xy = m
y = x - 60
m = x(x-60)
m = x^2 - 60x
m' = 2x - 60
x = 30 , y = -30

but y can't be negative...so? :confused:
 
But:
You should have
x+y=60 xy(max.)
Not x-y!
 
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