Lagrange Multipliers - Multivariable Calculus

1. A particle moves along the curve $x^2 + 2y^2 = 6$. At which point on this curve is the function $h(x,y) = x^2y$ maximized?
   Answer: $(2, 1)$ Using Lagrange multipliers with constraint $g(x,y) = x^2 + 2y^2 - 6 = 0$. We need $\nabla h = \lambda \nabla g$, giving $(2xy, x^2) = \lambda(2x, 4y)$. This yields $2xy = 2\lambda x$ and $x^2 = 4\lambda y$. From the first equation (assuming $x \neq 0$): $y = \lambda$. Substituting into the second: $x^2 = 4y^2$, so $x = \pm 2y$. Using the constraint $x^2 + 2y^2 = 6$: when $x = 2y$, we get $4y^2 + 2y^2 = 6$, so $6y^2 = 6$, giving $y = \pm 1$. When $x = -2y$, we get the same $y$ values. This gives us critical points: $(2,1)$, $(2,-1)$, $(-2,1)$, $(-2,-1)$. We also need to check the case $x = 0$: from the constraint, $2y^2 = 6$, so $y = \pm\sqrt{3}$, giving points $(0, \sqrt{3})$ and $(0, -\sqrt{3})$. Evaluating $h(x,y) = x^2y$: At $(2,1)$: $h = 4(1) = 4$. At $(2,-1)$: $h = 4(-1) = -4$. At $(-2,1)$: $h = 4(1) = 4$. At $(-2,-1)$: $h = 4(-1) = -4$. At $(0, \pm\sqrt{3})$: $h = 0$. The maximum value is 4, achieved at $(2,1)$ and $(-2,1)$. Since only $(2,1)$ appears in the options, this is the answer. Choice B doesn't satisfy the constraint. Choice C gives a minimum. Choice D gives $h = 0$.
2. Find the maximum value of the function $f(x, y) = 4x + 3y$ subject to the constraint given by the ellipse $\frac{x^2}{18} + \frac{y^2}{32} = 1$.
   Answer: 24 Let $g(x,y) = \frac{x^2}{18} + \frac{y^2}{32} - 1$. We set up the Lagrange multiplier system $\nabla f = \lambda \nabla g$. This gives $\langle 4, 3 \rangle = \lambda \langle \frac{x}{9}, \frac{y}{16} \rangle$. We have the equations $4 = \lambda \frac{x}{9}$ and $3 = \lambda \frac{y}{16}$. Solving for $x$ and $y$ gives $x = \frac{36}{\lambda}$ and $y = \frac{48}{\lambda}$. Substituting these into the constraint equation: $\frac{(36/\lambda)^2}{18} + \frac{(48/\lambda)^2}{32} = 1$. This simplifies to $\frac{1296}{18\lambda^2} + \frac{2304}{32\lambda^2} = 1$, which becomes $\frac{72}{\lambda^2} + \frac{72}{\lambda^2} = 1$, so $\frac{144}{\lambda^2} = 1$. Thus, $\lambda^2 = 144$, and $\lambda = \pm 12$. For $\lambda = 12$, we get $(x,y) = (3,4)$, and $f(3,4) = 4(3) + 3(4) = 24$. For $\lambda = -12$, we get $(x,y) = (-3,-4)$, and $f(-3,-4) = -24$. The maximum value is 24.
3. A manufacturer's production is modeled by the Cobb-Douglas function $P(K, L) = 100 K^{1/4} L^{3/4}$, where $K$ is units of capital and $L$ is units of labor. The budget for production is $120,000, with capital costing $300 per unit and labor costing $100 per unit. What is the maximum production level $P$?
   Answer: $30000\sqrt{3}$ This is a constrained optimization problem where you need to maximize a production function subject to a budget constraint. When you see a Cobb-Douglas function with a linear budget constraint, think Lagrange multipliers or the method of substitution. Set up the constraint equation: $300K + 100L = 120,000$, which simplifies to $3K + L = 1200$. From this, $L = 1200 - 3K$. Substitute into the production function: $P(K) = 100K^{1/4}(1200-3K)^{3/4}$. To find the maximum, take the derivative and set it equal to zero: $P'(K) = 100 \left[\frac{1}{4}K^{-3/4}(1200-3K)^{3/4} + K^{1/4} \cdot \frac{3}{4}(1200-3K)^{-1/4} \cdot (-3)\right] = 0$ Factoring out common terms and simplifying: $\frac{1200-3K}{4K} = \frac{9}{4}$ Solving: $1200-3K = 9K$, so $K = 100$ and $L = 900$. The maximum production is: $P(100,900) = 100 \cdot 100^{1/4} \cdot 900^{3/4} = 100 \cdot \sqrt[4]{100} \cdot \sqrt[4]{900^3}$ Since $100^{1/4} = \sqrt{10}$ and $900^{3/4} = (30^2)^{3/4} = 30^{3/2} = 30\sqrt{30} = 30\sqrt{30}$: $P = 100 \cdot \sqrt{10} \cdot 30\sqrt{30} = 30000\sqrt{3}$ Option A (30,000) ignores the irrational component. Option B (40,000) likely comes from computational errors. Option C ($25\sqrt{3}$) has the right irrational form but wrong magnitude, probably from early algebraic mistakes. Study tip: In Cobb-Douglas optimization problems, always double-check your arithmetic with radicals—the final answer often involves square roots that students accidentally drop or miscalculate.
4. A consumer's utility from consuming $x$ units of good A and $y$ units of good B is given by $U(x,y) = xy$. The price of good A is $2 per unit and the price of good B is $5 per unit. The consumer's budget is $100. Using the method of Lagrange multipliers, the value of the multiplier $\lambda$ represents the marginal utility of money. What is the approximate increase in maximum utility if the consumer's budget increases from $100 to $101?
   Answer: 5 We maximize $U(x,y)=xy$ subject to $g(x,y)=2x+5y=100$. The Lagrange system is $\nabla U = \lambda \nabla g$, which is $\langle y, x \rangle = \lambda \langle 2, 5 \rangle$. This yields $y=2\lambda$ and $x=5\lambda$. Substituting into the budget constraint: $2(5\lambda) + 5(2\lambda) = 100 \implies 10\lambda + 10\lambda = 100 \implies 20\lambda = 100 \implies \lambda = 5$. The value of $\lambda$ approximates the change in the optimal value of $U$ for a one-unit increase in the constraint constant. Therefore, if the budget increases by $1, the maximum utility will increase by approximately $\lambda = 5$.
5. Let $P$ be a point on the ellipse $x^2 + 2y^2 = 6$ where the function $f(x,y) = x^2y$ attains its maximum value. Which of the following statements provides the correct geometric interpretation of the relationship between the ellipse and the level curve of $f$ that passes through $P$?
   Answer: The ellipse and the level curve are tangent at $P$. The method of Lagrange multipliers is based on the principle that at an constrained extremum $P$, the gradient of the function $f$ must be parallel to the gradient of the constraint function $g$. That is, $\nabla f(P) = \lambda \nabla g(P)$. The gradient vector $\nabla f$ is orthogonal to the level curves of $f$, and $\nabla g$ is orthogonal to the level curve of $g$ (which is the constraint curve). If the gradient vectors are parallel, their corresponding level curves must be tangent at the point $P$.
6. What is the absolute minimum value of the function $f(x,y) = x^2 - y^2$ subject to the constraint $x^2 + y^2 = 4$?
   Answer: -4 Let $g(x,y) = x^2+y^2-4=0$. The Lagrange system is $\nabla f = \lambda \nabla g$, which gives $\langle 2x, -2y \rangle = \lambda \langle 2x, 2y \rangle$. This results in two equations: (1) $2x = 2\lambda x$ and (2) $-2y = 2\lambda y$. From (1), $2x(1-\lambda)=0$, so either $x=0$ or $\lambda=1$. From (2), $2y(1+\lambda)=0$, so either $y=0$ or $\lambda=-1$. We examine the cases: Case 1: $x=0$. The constraint gives $y^2=4$, so $y=\pm 2$. The points are $(0, 2)$ and $(0, -2)$. For both points, $f(x,y) = 0^2 - (\pm 2)^2 = -4$. Case 2: $y=0$. The constraint gives $x^2=4$, so $x=\pm 2$. The points are $(2, 0)$ and $(-2, 0)$. For both points, $f(x,y) = (\pm 2)^2 - 0^2 = 4$. We cannot have $\lambda=1$ and $\lambda=-1$ simultaneously. Comparing the function values, the absolute minimum is -4.
7. Let $P=(x,y,z)$ be the point on the sphere $x^2 + y^2 + z^2 = 84$ that is closest to the point $(1, 2, 3)$. What is the value of the sum of the coordinates, $x+y+z$?
   Answer: $6\sqrt{6}$ We want to minimize the distance, which is equivalent to minimizing the squared distance function $f(x,y,z) = (x-1)^2 + (y-2)^2 + (z-3)^2$ subject to the constraint $g(x,y,z) = x^2 + y^2 + z^2 = 84$. The Lagrange system is $\nabla f = \lambda \nabla g$, which gives $\langle 2(x-1), 2(y-2), 2(z-3) \rangle = \lambda \langle 2x, 2y, 2z \rangle$. This leads to $x-1=\lambda x$, $y-2=\lambda y$, and $z-3=\lambda z$. Solving for $x, y, z$ gives $x=\frac{1}{1-\lambda}$, $y=\frac{2}{1-\lambda}$, $z=\frac{3}{1-\lambda}$. This implies $y=2x$ and $z=3x$. Substituting into the constraint: $x^2 + (2x)^2 + (3x)^2 = 84 \implies 14x^2 = 84 \implies x^2 = 6 \implies x = \pm\sqrt{6}$. The two candidate points are $(\sqrt{6}, 2\sqrt{6}, 3\sqrt{6})$ and $(-\sqrt{6}, -2\sqrt{6}, -3\sqrt{6})$. The point $(\sqrt{6}, 2\sqrt{6}, 3\sqrt{6})$ is closer to $(1,2,3)$. The sum of its coordinates is $\sqrt{6} + 2\sqrt{6} + 3\sqrt{6} = 6\sqrt{6}$.
8. The method of Lagrange multipliers is used to find the extrema of a differentiable function $f$ subject to a differentiable constraint $g(x,y,z)=c$. The method is guaranteed to identify all absolute maximum and minimum values of $f$ under which of the following conditions?
   Answer: The set of points satisfying $g(x,y,z)=c$ is closed and bounded, and $\nabla g \neq \mathbf{0}$ on this set. When applying Lagrange multipliers to find extrema of a function $f$ subject to constraint $g(x,y,z)=c$, you're looking for points where $\nabla f = \lambda \nabla g$. However, this method alone doesn't guarantee you'll find all absolute extrema—you need additional conditions to ensure completeness. The correct answer is C because it combines two essential requirements. First, the constraint set must be closed and bounded (compact), which guarantees by the Extreme Value Theorem that continuous function $f$ will attain its absolute maximum and minimum on this set. Second, $\nabla g \neq \mathbf{0}$ everywhere on the constraint set ensures the constraint qualifies as a "regular" curve/surface where the Lagrange multiplier method is valid—this prevents degenerate cases where the method might miss critical points. Option A gives you the regularity condition but lacks compactness, so extrema might not exist (imagine optimizing on an unbounded curve). Option B provides compactness but without $\nabla g \neq \mathbf{0}$, you could have singular points where Lagrange multipliers fail—like trying to optimize at a cusp or corner where the constraint isn't smooth. Option D incorrectly suggests that specific function types (linear $f$, quadratic $g$) alone guarantee the method works, but the geometric properties of the constraint set matter more than the algebraic forms. Study tip: Remember that Lagrange multipliers need both analytical conditions ($\nabla g \neq \mathbf{0}$) and topological conditions (closed, bounded constraint set) to guarantee finding all absolute extrema. Always check both requirements.
9. An open-topped rectangular box with a square base is to be constructed with a volume of exactly 32 cubic meters. What is the minimum possible surface area of the material needed for its construction?
   Answer: 48 m$^2$ Let the square base have side length $x$ and the box have height $y$. The volume is $V = x^2y = 32$. This is the constraint. The surface area of the open-topped box is $S = x^2 + 4xy$. We want to minimize $S$. Let $f(x,y) = x^2+4xy$ and $g(x,y) = x^2y - 32$. The system $\nabla f = \lambda \nabla g$ gives $\langle 2x+4y, 4x \rangle = \lambda \langle 2xy, x^2 \rangle$. From the second component, $4x = \lambda x^2$, which implies $\lambda = 4/x$ (since $x>0$). Substituting into the first component equation: $2x+4y = (4/x)(2xy) = 8y$. This simplifies to $2x = 4y$, or $x=2y$. Substitute this into the volume constraint: $(2y)^2 y = 32 \implies 4y^3 = 32 \implies y^3 = 8 \implies y=2$. Then $x = 2(2) = 4$. The minimum surface area is $S = (4)^2 + 4(4)(2) = 16 + 32 = 48$ m$^2$.
10. Find the maximum value of the function $f(x,y,z) = x - 2y + 2z$ for points on the curve formed by the intersection of the cylinder $x^2 + y^2 = 1$ and the plane $x+y+z=1$.
    Answer: $2 + \sqrt{17}$ This is a constrained optimization problem where you need to find the maximum of a linear function subject to two constraints simultaneously. When you see this setup, think Lagrange multipliers - you're optimizing over the intersection of two surfaces. The curve is defined by two constraints: $g_1(x,y,z) = x^2 + y^2 - 1 = 0$ (cylinder) and $g_2(x,y,z) = x + y + z - 1 = 0$ (plane). Using Lagrange multipliers, you need $\nabla f = \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2$. This gives you: $\nabla f = (1, -2, 2)$, $\nabla g_1 = (2x, 2y, 0)$, and $\nabla g_2 = (1, 1, 1)$. Setting up the system: - $1 = 2\lambda_1 x + \lambda_2$ - $-2 = 2\lambda_1 y + \lambda_2$ - $2 = \lambda_2$ From the third equation, $\lambda_2 = 2$. Substituting back: $2\lambda_1 x = -1$ and $2\lambda_1 y = -4$, so $x = -\frac{1}{2\lambda_1}$ and $y = -\frac{2}{\lambda_1}$. Since these points must satisfy $x^2 + y^2 = 1$, you get $\lambda_1 = \pm\frac{\sqrt{17}}{2}$. For maximum value, use $\lambda_1 = -\frac{\sqrt{17}}{2}$, giving $(x,y) = \left(\frac{1}{\sqrt{17}}, \frac{4}{\sqrt{17}}\right)$. From the plane constraint, $z = 1 - x - y = 1 - \frac{5}{\sqrt{17}}$. Thus $f = \frac{1}{\sqrt{17}} - \frac{8}{\sqrt{17}} + 2 - \frac{10}{\sqrt{17}} = 2 + \sqrt{17}$. Choice (A) 3 likely comes from ignoring one constraint. Choice (B) $2 - \sqrt{17}$ is the minimum value. Choice (D) 1 might result from computational errors. Remember: constrained optimization with multiple constraints requires Lagrange multipliers, and always check whether you found a maximum or minimum by examining the sign of your multipliers.
11. Consider the optimization problem: maximize $f(x,y) = xy$ subject to $x^2 + y^2 = 8$. Which statement about the critical points is correct?
    Answer: There are exactly four critical points, with two giving local maxima with function value 4 Using Lagrange multipliers: $\nabla f = \lambda \nabla g$ where $g(x,y) = x^2 + y^2 - 8$. This gives $y = 2\lambda x$ and $x = 2\lambda y$. Substituting the first into the second: $x = 2\lambda(2\lambda x) = 4\lambda^2 x$. For $x \neq 0$, we get $1 = 4\lambda^2$, so $\lambda = \pm\frac{1}{2}$. When $\lambda = \frac{1}{2}$: $y = x$, and from the constraint $2x^2 = 8$, so $x = \pm 2$, giving points $(2,2)$ and $(-2,-2)$ with $f = 4$. When $\lambda = -\frac{1}{2}$: $y = -x$, giving points $(2,-2)$ and $(-2,2)$ with $f = -4$. The points with $f = 4$ are local maxima. Choice A misses half the critical points. Choice C uses incorrect constraint radius. Choice D both misses critical points and has wrong function value.
12. Find the maximum value of $f(x,y,z) = x + 2y + 3z$ subject to the constraints $x^2 + y^2 + z^2 = 14$ and $x + y + z = 0$.
    Answer: $2\sqrt{7}$ This requires two Lagrange multipliers since we have two constraints. Let $g_1(x,y,z) = x^2 + y^2 + z^2 - 14$ and $g_2(x,y,z) = x + y + z$. The conditions are: $\nabla f = \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2$, giving us $1 = 2\lambda_1 x + \lambda_2$, $2 = 2\lambda_1 y + \lambda_2$, $3 = 2\lambda_1 z + \lambda_2$. From these: $x = \frac{1-\lambda_2}{2\lambda_1}$, $y = \frac{2-\lambda_2}{2\lambda_1}$, $z = \frac{3-\lambda_2}{2\lambda_1}$. Using the constraint $x + y + z = 0$: $\frac{(1-\lambda_2) + (2-\lambda_2) + (3-\lambda_2)}{2\lambda_1} = 0$, so $6 - 3\lambda_2 = 0$, giving $\lambda_2 = 2$. Thus $x = \frac{-1}{2\lambda_1}$, $y = 0$, $z = \frac{1}{2\lambda_1}$. Using $x^2 + y^2 + z^2 = 14$: $\frac{1}{4\lambda_1^2} + 0 + \frac{1}{4\lambda_1^2} = 14$, so $\frac{1}{2\lambda_1^2} = 14$, giving $\lambda_1^2 = \frac{1}{28}$, so $\lambda_1 = \pm\frac{1}{2\sqrt{7}}$. For maximum, we want $\lambda_1 = -\frac{1}{2\sqrt{7}}$, giving $x = \sqrt{7}$, $y = 0$, $z = -\sqrt{7}$. The maximum value is $\sqrt{7} + 0 + 3(-\sqrt{7}) = \sqrt{7} - 3\sqrt{7} = -2\sqrt{7}$. Wait, this is negative. For maximum, we want $\lambda_1 = \frac{1}{2\sqrt{7}}$, giving $x = -\sqrt{7}$, $y = 0$, $z = \sqrt{7}$. The maximum value is $-\sqrt{7} + 0 + 3\sqrt{7} = 2\sqrt{7}$. Choice A comes from computational errors. Choice C results from incorrect constraint handling. Choice D represents incorrect combination of the constraint values.
13. When attempting to find the extrema of $f(x,y) = x+y$ subject to the constraint $g(x,y) = x^3 - y^2 = 0$, the method of Lagrange multipliers fails to identify the point $(0,0)$ as a potential extremum. What is the reason for this failure?
    Answer: The gradient of the constraint function, $\nabla g$, is the zero vector at $(0,0)$. When using Lagrange multipliers to find extrema of $f(x,y)$ subject to constraint $g(x,y) = 0$, you're looking for points where $\nabla f = \lambda \nabla g$ for some scalar $\lambda$. This method requires that $\nabla g \neq \mathbf{0}$ at potential extrema—this is called the regularity condition. Let's examine what happens at $(0,0)$. For $g(x,y) = x^3 - y^2$, we have $\nabla g = (3x^2, -2y)$. At $(0,0)$, this gives us $\nabla g(0,0) = (0,0) = \mathbf{0}$. Since the gradient of the constraint function is zero, the Lagrange multiplier equation $\nabla f = \lambda \nabla g$ becomes $\nabla f = \lambda \mathbf{0} = \mathbf{0}$. But $\nabla f = (1,1) \neq \mathbf{0}$, so no $\lambda$ can satisfy this equation. The method fails because the regularity condition is violated. Answer A is incorrect because $f(x,y) = x + y$ is clearly differentiable everywhere, including at $(0,0)$. Answer B misses the point—while the constraint curve $x^3 = y^2$ is indeed unbounded, this isn't why Lagrange multipliers fails at this specific point. Answer D is wrong because $\nabla f = (1,1)$ is never zero. Remember: Lagrange multipliers can fail when $\nabla g = \mathbf{0}$ at a point on the constraint curve. Always check the regularity condition, and when it fails, examine those points separately as potential extrema.
14. What is the minimum value of the function $S(x,y) = 2x+8y$ for points $(x,y)$ in the first quadrant on the hyperbola $xy=9$?
    Answer: 24 We want to minimize $f(x,y) = 2x+8y$ subject to $g(x,y) = xy-9=0$ with $x,y>0$. The Lagrange system is $\nabla f = \lambda \nabla g$, which gives $\langle 2, 8 \rangle = \lambda \langle y, x \rangle$. This yields the equations $2=\lambda y$ and $8=\lambda x$. Since $x,y>0$, $\lambda$ must be non-zero. From these equations, $y = 2/\lambda$ and $x=8/\lambda$, which implies $x=4y$. Substituting this into the constraint: $(4y)y = 9 \implies 4y^2 = 9 \implies y=3/2$ (since we are in the first quadrant). Then $x=4(3/2)=6$. The minimum value occurs at $(6, 3/2)$ and is $S(6, 3/2) = 2(6) + 8(3/2) = 12 + 12 = 24$. Distractor A comes from an error in applying the AM-GM inequality. Distractors C and D come from testing simple integer points on the hyperbola like $(9,1)$ and $(3,3)$.

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