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# Minimization problem?

Find the least value of (32x^2+11y^2+25z^2)/(xy+yz+zx) for x,y,z all positive

@John - https://imgur.com/tyrR8gb

@atsuo - thankyou - correct answer but calculus not what I was looking for. It is solvable using only basic algebra. Of course Wolfram solves original problem directly.

### 1 Answer

- atsuoLv 61 month ago
f(x,y,z) = (32x^2 + 11y^2 + 25z^2)/(xy + yz + zx) for x,y,z > 0.

Let y = mx and z = nx. So m,n > 0 and

f = (32x^2 + 11y^2 + 25z^2)/(xy + yz + zx)

= (32x^2 + 11m^2x^2 + 25n^2x^2)/(mx^2 + mnx^2 + nx^2)

= (32 + 11m^2 + 25n^2)/(m + mn +n) ---(#1)

(#1) is continuous in the 1st quadrant of the mn-plane. If f has the least value then ∂f/∂m = ∂f/∂n = 0 at this point.

∂f/∂m = [22m(m + mn + n)-(32 + 11m^2 + 25n^2)(1 + n)]/(m + mn + n)^2 = 0

22m(m + mn + n) - (32 + 11m^2 + 25n^2)(1 + n) = 0

22m^2 + 22m^2n + 22mn - 32 - 11m^2 - 25n^2 - 32n -11m^2n - 25n^3 = 0

11m^2 + 11m^2n + 22mn - 25n^3 - 25n^2 - 32n - 32 = 0 ---(#2)

∂f/∂n = [50n(m + mn + n) - (32 + 11m^2 + 25n^2)(1 + m)]/(m + mn + n)^2 = 0

50n(m + mn + n) - (32 + 11m^2 + 25n^2)(1 + m) = 0

50mn + 50mn^2 + 50n^2 - 32 - 11m^2 - 25n^2 - 32m - 11m^3 - 25mn^2 = 0

25n^2 + 25mn^2 + 50mn - 11m^3 - 11m^2 - 32m - 32 = 0 ---(#3)

(#2) and (#3) are simultaneous equations of m and n, but they are too complicated. So I asked Wolfram|alpha and it said (m,n) = (2,6/5) is the only solution in the 1st quadrant.

For example, let x = 5. So y = 10 and z = 6.

f(5,10,6) = (32*25 + 11*100 + 25*36) / (50 + 60 + 30) = 20 <--- the answer