find the max and min values of f(x,y)=x^2+4y^2-x+2y, bounded by the ellipse x^2+

Question

find the max and min values of f(x,y)=x^2+4y^2-x+2y, bounded by the ellipse x^2+4y^2=1

Explanation / Answer

1) First, find the critical points of f inside (or on) S by setting f_x = 0 and f_y = 0. f_x = 2x = 0 and f_y = 2y = 0 ==> (x, y) = (0, 0), which is inside S. Now, we find the critical values on the boundary of S. Let f(x,y) = x^2 + y^2, subject to g(x, y) = (x - 1)^2 + 4y^2 = 4. So, ?f = ??g ==> = ? ==> 2x = 2?(x - 1) and 2y = 8?y ==> x = ?(x - 1) and (4? - 1)y = 0. If y = 0, then (x - 1)^2 = 4 ==> x = -1 or 3. If ? = 1/4, then we find that (x, y) = (-1/3, ±v5/3). Testing all of these points: f(0, 0) = 0 8yz = 2?x, 8xz = 2?y, 8xy = 8?z. So, 8xyz = 2?x^2 = 2?y^2 = 8?z^2 ==> x^2 = y^2 = 4z^2, since ? = 0 ==> At least one of x,y,z = 0. Substituting this into g yields x^2 + x^2 + x^2 = 4 ==> x = ±2/v3. Since we want x, y, z > 0, we have the critical point (x, y, z) = (2/v3, 2/v3, 1/v3). This yields maximal volume V = 8xyz = 32/(3v3). -------------------------- 3) We want to mimimize the square of the distance D = (x - 1)^2 + (y - 1)^2 + (z - 100)^2 subject to z = 100 - x^2 - y^2. For a change of pace, let's minimize D = (x - 1)^2 + (y - 1)^2 + [(100 - x^2 - y^2) - 100]^2 ...= (x - 1)^2 + (y - 1)^2 + (x^2 + y^2)^2. D_x = 2(x - 1) + 4x(x^2 + y^2), and D_y = 2(y - 1) + 4y(x^2 + y^2). Set these equal to 0: 2(x - 1) + 4x(x^2 + y^2) = 0, 2(y - 1) + 4y(x^2 + y^2) = 0. ==> -4x(x^2 + y^2) = 2y(x - 1) = 2x(y - 1) ==> x = y. Substitute this into either equation: 2(x - 1) + 4x(x^2 + x^2) = 0 ==> 4x^3 + x - 1 = 0 ==> (2x - 1)(2x^2 + x + 1) = 0 ==> x = 1/2 (other roots are not real) So, the minimal distance occurs at (1/2, 1/2) ==> d = vD = v3/2.

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

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