# Optimization Problem?

Find two positive real numbers such that they sum to 102 and the product of the first times the square of the second is a maximum.

### 4 Answers

- PhilipLv 62 months ago
Put numbers = x & (102-x).

Put f(x) = x(102-x)^2 = x(x-102)^2 = x[x^2 -204x + a^2], where a = 102.

Then f(x) = x^3 -204x^2 + a^2x.

f'(x) = 3x^2 - 408x + a^2.

f''(x) = 6x - 408 = 6(x-68).

Extremums occur where f'(x) = 0, ie., where x^2 -136x + (1/3)[3*34]^2 = 0, ie., where

g(x) = x^2 - 136x + 3(34)^2 = 0...(1). for g(x) = 0, 2x = 136(+/-)D, where D^2 = (136)^2 - 4*3*(34)^2 = 16(34)^2 - 12(34)^2 = 4(34)^2. Then D =2*34 =68 and

2x = 136(+/-)68 = 68[2(+/-)1]. Then x = 34(1 or 3) = 34...(2) or x = 102...(3). Clearly,

root x = 102 is invalid. Therefore x = 34.

At x = 34, f''(x) = 6*34 - 408 = -(408-204) = -204, < 0 ---> extremum at x = 34 is a

maximum.

f(34) = 34(102-34)^2 = 34[3*34 - 34] = 34[2*34] = 2*17[4*17] = 8*(17)^2 = 8*(289) =

8*(290-1) = 2320-8 = 2312.

The 2 numbers are 34 and 68 such that f(34) = maximum = 2312 = 34*68.

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- davidLv 72 months ago
n = 1st number

102 - n = 2nd number

f(n) = n(102-n)^2 <<< derivative of this

f '(n) = 3(n^2 - 136 n + 3468)

n^2 - 136 n + 3468 = 0

(n-102)(n-34) = 0

n = 34 and n = 102<< not possible, 2ns number would be 0

n = 34 ... 102 - n = 68 <<< answers

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- sepiaLv 72 months ago
Find two positive real numbers such that they sum to 102

and the product of the first times the square of the second is a maximum.

x + y = 102

x(y^2) = 101^2

(102 - y)y^2 = 101^2

y^3 - 102y^2 + 101^2 = 0

y = 101 and x = 1

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- KrishnamurthyLv 72 months ago
Find two positive real numbers such that they sum to 102

and the product of the first times the square of the second is a maximum.

x + y = 102

x(y^2) = 51^3

(102 - y)y^2 = 51^3

y^3 - 102y^2 + 51^3 = 0

y = 51

The two positive real numbers are 51 and 51

- PhilipLv 62 months agoReport
Krisnamurthy, would you please be so kind as to check my solution for errors?

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