Lagrange Multipliers - Calculus 3

1. Consider the constrained optimization problem: minimize $f(x,y) = x^2 + y^2$ subject to $(x-1)^2 + (y-1)^2 = 2$. A student sets up the Lagrangian and finds the condition $2x = 2\lambda(x-1)$ and $2y = 2\lambda(y-1)$. If they conclude that $x = y$ at the critical points, which additional step is necessary to find the minimum value of $f$?
   Answer: Substitute $x = y$ into the constraint to get $2(x-1)^2 = 2$, giving $x = y = 0$ or $x = y = 2$, then evaluate $f$ at both When approaching constrained optimization with Lagrange multipliers, you need to systematically work through the critical point conditions to find all candidates, then evaluate the objective function at each to determine the optimum. The student correctly derived that $x = y$ from the Lagrangian conditions. Now you must find the actual coordinates of these critical points. Option A does exactly this: substituting $x = y$ into the constraint $(x-1)^2 + (y-1)^2 = 2$ gives $2(x-1)^2 = 2$, which simplifies to $(x-1)^2 = 1$. This yields $x-1 = \pm 1$, so $x = 0$ or $x = 2$, giving critical points $(0,0)$ and $(2,2)$. Evaluating $f(x,y) = x^2 + y^2$ at both: $f(0,0) = 0$ and $f(2,2) = 8$. Since you're minimizing, the minimum value is 0. Option B is incorrect because checking $\lambda \neq 1$ isn't the next logical step—you already have the key relationship $x = y$ and should use it immediately. Option C involves second-order conditions, but for a continuous function on a compact constraint set (a circle), you can simply evaluate $f$ at all critical points to find the minimum. Option D suggests an alternative method entirely, which isn't necessary when the Lagrangian approach is already set up correctly. Study tip: In Lagrange multiplier problems, once you've simplified the gradient conditions, immediately substitute back into the constraint to find specific coordinates, then evaluate your objective function at each point.
2. A rectangular box with square base is to be constructed with a total surface area of $150$ square units. If the base has side length $s$ and height $h$, and we want to maximize the volume $V = s^2h$ subject to the constraint $s^2 + 4sh = 150$, what is the ratio $\frac{h}{s}$ at the optimal point?
   Answer: $\frac{h}{s} = \frac{1}{2}$ The Lagrangian is $L(s,h,\lambda) = s^2h - \lambda(s^2 + 4sh - 150)$. Taking partial derivatives: $\frac{\partial L}{\partial s} = 2sh - \lambda(2s + 4h) = 0$ and $\frac{\partial L}{\partial h} = s^2 - \lambda(4s) = 0$. From the second equation: $\lambda = \frac{s^2}{4s} = \frac{s}{4}$. Substituting into the first equation: $2sh - \frac{s}{4}(2s + 4h) = 0$, which gives $2sh - \frac{s^2}{2} - sh = 0$, so $sh - \frac{s^2}{2} = 0$. Since $s > 0$, we can divide by $s$: $h - \frac{s}{2} = 0$, giving $h = \frac{s}{2}$. Therefore $\frac{h}{s} = \frac{1}{2}$. We can verify this satisfies the constraint: $s^2 + 4s \cdot \frac{s}{2} = s^2 + 2s^2 = 3s^2 = 150$, so $s^2 = 50$ and $h = \frac{s}{2} = \frac{\sqrt{50}}{2}$.
3. Consider maximizing $f(x,y,z) = xyz$ subject to $x^2 + y^2 + z^2 = 3$. A student finds that at the critical points, $x^2 = y^2 = z^2 = 1$. However, when checking the constraint qualification, they discover an issue. What is the primary concern with this approach?
   Answer: There are multiple critical points with different function values, and the student must determine which gives the global maximum Using Lagrange multipliers: $\nabla f = \lambda \nabla g$ gives $(yz, xz, xy) = \lambda(2x, 2y, 2z)$. This leads to $x^2 = y^2 = z^2 = 1$, giving eight critical points: $(\pm 1, \pm 1, \pm 1)$. Evaluating $f$ at these points: $f(1,1,1) = f(-1,-1,1) = f(-1,1,-1) = f(1,-1,-1) = 1$ and $f(-1,-1,-1) = f(1,1,-1) = f(1,-1,1) = f(-1,1,1) = -1$. The constraint is smooth everywhere since $\nabla g = (2x,2y,2z) \neq \mathbf{0}$ on the sphere. Both $f$ and $g$ are differentiable everywhere. The issue is that the student found the correct critical point condition but must evaluate all eight points to determine that the maximum value is $1$.
4. Find the minimum distance from the origin to the curve defined by the intersection of the cylinder $x^2 + y^2 = 4$ and the plane $x + y + z = 6$. Using Lagrange multipliers with two constraints, which of the following represents the correct system of equations?
   Answer: $2x = 2\lambda_1 x + \lambda_2, 2y = 2\lambda_1 y + \lambda_2, 2z = \lambda_2$ with constraints $x^2 + y^2 = 4$ and $x + y + z = 6$ We minimize $f(x,y,z) = x^2 + y^2 + z^2$ subject to $g_1(x,y,z) = x^2 + y^2 - 4 = 0$ and $g_2(x,y,z) = x + y + z - 6 = 0$. The gradients are: $\nabla f = (2x, 2y, 2z)$, $\nabla g_1 = (2x, 2y, 0)$, $\nabla g_2 = (1, 1, 1)$. The Lagrange condition is $\nabla f = \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2$, giving: $2x = \lambda_1(2x) + \lambda_2(1) = 2\lambda_1 x + \lambda_2$, $2y = \lambda_1(2y) + \lambda_2(1) = 2\lambda_1 y + \lambda_2$, $2z = \lambda_1(0) + \lambda_2(1) = \lambda_2$. Choice A has incorrect coefficients (missing factor of 2 in gradients). Choice B has an error in the gradient of the constraint. Choice C has the gradients backwards. Choice D correctly represents the system.
5. A temperature function $T(x,y,z) = 100 - x^2 - y^2 - z^2$ is defined in a region, and we want to find the maximum temperature on the intersection curve of the surfaces $x + y + z = 3$ and $x^2 + y^2 = 2$. Using the method of Lagrange multipliers with two constraints, how many critical points should we expect to find?
   Answer: Exactly 2 critical points, corresponding to the highest and lowest points on the intersection curve The intersection of $x + y + z = 3$ (a plane) and $x^2 + y^2 = 2$ (a circular cylinder) is a circle in 3D space. We can parameterize this curve: let $x = \sqrt{2}\cos t$ and $y = \sqrt{2}\sin t$, then $z = 3 - x - y = 3 - \sqrt{2}(\cos t + \sin t)$. The temperature along this curve is $T(t) = 100 - 2 - (3 - \sqrt{2}(\cos t + \sin t))^2 = 98 - (3 - \sqrt{2}(\cos t + \sin t))^2$. To maximize this, we minimize $(3 - \sqrt{2}(\cos t + \sin t))^2$, which occurs when $\cos t + \sin t$ is maximized or minimized. Since $\cos t + \sin t = \sqrt{2}\sin(t + \pi/4)$, the extrema occur when $\sin(t + \pi/4) = \pm 1$, giving exactly 2 critical points. These correspond to the points on the intersection circle that are farthest from and closest to the origin (since $T$ decreases with distance from origin). Choice A is wrong because the plane intersects the cylinder in a full circle, not tangentially. Choice C misapplies the constraint counting. Choice D is wrong because symmetry doesn't create infinitely many critical points.

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