find two numbers whose sum is 36 if the product of one by the square of the other is a maximum
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Answer:
The sum of two non-negative numbers is 36. What are the numbers if the product of one number and the square root of the other is to be as large as possible?
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This is an optimization problem.
You want to maximize a*sqrt(b) with the constraint that a+b = 36
Well, I will first assume that you admit every real number, I mean, that the problem is not restricted to natural numbers…
And I will also ignore, only at the beginning, that the numbers must be both non-negative.
Well, optimization in real numbers is usually solved with derivatives.
We have:
a+b = 36
→ a = 36–b
With this we have now a function of one variable to maximize:
a*sqrt(b) = (36-b) * sqrt(b)
Let
f(x) = (36-x) * sqrt(x) = (36-x) * x^(1/2)
To find the maximum we compute the derivative:
f’(x) = -1*sqrt(x) + (36-x) * (1/2) * x^(-1/2) =
= -sqrt(x) + (1/2) * (36-x) / sqrt(x)
We are searching for values where f’(x) = 0
-sqrt(x) + (1/2) * (36-x) / sqrt(x) = 0
→ (multipying by 2sqrt(x) →
→ -2x + (36-x) = 0
→ -3x + 36 = 0
→ x = 12
For b = 12, a = 24 … both positive numbers and also natural numbers.
The maximum product is : 24*sqrt(12) = 83.13843876…
We can check that is higher than 20*sqrt(16) = 20*4 = 80
Yes, it is.
And that is also higher than 27*sqrt(9) = 27*3 = 81
Yes, it is.
And that is also higher than 11*sqrt(25) = 11*5 = 55