Find the minimum value of (2x + 3y)(x + 3z)(y + 2z) where x; y; z > 0 and xyz =

Question

Find the minimum value of (2x + 3y)(x + 3z)(y + 2z) where x; y; z > 0 and xyz = 1.

Explanation / Answer

Since xyz = 1, we can rewrite this as P = 2x^2 y + 4x^2 z + 3xy^2 + 12xyz + 12xz^2 + 9y^2 z + 18yz^2 ...= 2x/z + 4x/y + 3y/z + 12 * 1 + 12z/y + 9y/x + 18z/x ...= 2x/z + 18z/x + 4x/y + 9y/x + 3y/z + 12z/y + 12 Let u = x/z, v = x/y and w = y/z. ==> P = 2u + 18/u + 4v + 9/v + 3w + 12/w + 12 Now, complete the square: P = 2(u + 9/u) + 4(v + 9/(4v)) + 3(w + 4/w) + 12 ...= 2(u - 6 + 9/u) + 4(v - 3 + 9/(4v)) + 3(w - 4 + 4/w) + 12 + 2 * 6 + 4 * 3 + 3 * 4 ...= 2(vu - 3/vu)^2 + 4(vv - 3/(2vv))^2 + 3(vw - 2/vw)^2 + 48. So, P has a minimum value of 48 if vu - 3/vu = 0, vv - 3/(2vv) = 0, and vw - 2/vw = 0 u = 3, v = 3/2, and w = 2. x = 3 * 6^(-1/3), y = 2 * 6^(-1/3), z = 6^(-1/3), by solving u = x/z, v = x/y and w = y/z, and xyz = 1 -------------- Hence, the minimum value is indeed 48.

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.