# Lagrange Multipliers Practice Problems with Answers > 12 Lagrange multiplier practice problems with answers: max and min on lines and circles, closest points, three variables and the meaning of lambda. Source: https://practicecalculus.com/practice-problems/lagrange-multipliers Updated: 2026-11-06 $$ \nabla f = \lambda\,\nabla g, \qquad g(x, y) = k $$ ## At a glance - **Problems:** 12, with answers and full solutions - **Level:** Calculus 3 - **You need:** [Partial derivatives](/practice-problems/partial-derivatives) and solving systems of equations - **Main idea:** At a constrained max or min, the gradient of $f$ is parallel to the gradient of the constraint - **Practice with new numbers:** [Lagrange multipliers lesson](/learn/calculus-3/partial-derivatives/lagrange-multipliers) ## Key takeaways - Write $\nabla f = \lambda\nabla g$ as one equation per variable, and add the constraint itself. - Solve by eliminating $\lambda$: divide or compare equations to get a relation between the variables. - Plug every candidate point into $f$. The largest value is the max and the smallest is the min. These **Lagrange multiplier practice problems** cover two-variable problems on lines and circles, closest-point problems, three-variable problems and what $\lambda$ means. Try each one first. The answers come first, then full solutions. The method, to find extreme values of $f$ subject to $g = k$: $$\nabla f = \lambda\,\nabla g, \qquad g(x, y) = k$$ Solve the system, then compare $f$ at every solution. For the geometry behind it, read the [Lagrange multipliers guide](/blog/lagrange-multipliers). ## Problems **Constraint on a line** 1. Maximize $f(x, y) = xy$ subject to $x + y = 10$. 2. Minimize $f(x, y) = x^2 + y^2$ subject to $x + y = 4$. 3. Find the point on the line $2x + y = 5$ closest to the origin. **Constraint on a circle** 4. Find the max and min of $f(x, y) = x + y$ on $x^2 + y^2 = 2$. 5. Find the max and min of $f(x, y) = xy$ on $x^2 + y^2 = 8$. 6. Find the max and min of $f(x, y) = x^2y$ on $x^2 + y^2 = 3$. 7. Find the max and min of $f(x, y) = 2x + 3y$ on $x^2 + y^2 = 13$. **Three variables and applications** 8. Maximize $f(x, y, z) = xyz$ subject to $x + y + z = 12$, with $x, y, z > 0$. 9. Find the point on the plane $x + 2y + 2z = 9$ closest to the origin. 10. A closed box has volume $8$. Find the dimensions with the least surface area. 11. A rectangle has perimeter $40$. Find its largest possible area. 12. In problem 1, find $\lambda$. Use it to estimate the new maximum if the constraint becomes $x + y = 11$, and compare with the exact value. ## Answers | Problem | Answer | | --- | --- | | 1 | Max $25$ at $(5, 5)$ | | 2 | Min $8$ at $(2, 2)$ | | 3 | $(2, 1)$, distance $\sqrt{5}$ | | 4 | Max $2$ at $(1, 1)$, min $-2$ at $(-1, -1)$ | | 5 | Max $4$ at $\pm(2, 2)$, min $-4$ at $\pm(2, -2)$ | | 6 | Max $2$ at $(\pm\sqrt{2}, 1)$, min $-2$ at $(\pm\sqrt{2}, -1)$ | | 7 | Max $13$ at $(2, 3)$, min $-13$ at $(-2, -3)$ | | 8 | Max $64$ at $(4, 4, 4)$ | | 9 | $(1, 2, 2)$, distance $3$ | | 10 | A $2\times 2\times 2$ cube, surface area $24$ | | 11 | $100$, a $10\times 10$ square | | 12 | $\lambda = 5$; estimate $30$, exact $30.25$ | ## Step-by-step solutions **1.** $\nabla f = \langle y, x\rangle$ and $\nabla g = \langle 1, 1\rangle$. So $y = \lambda$ and $x = \lambda$, which gives $x = y$. The constraint gives $x = y = 5$ and $f = 25$. **2.** $\langle 2x, 2y\rangle = \lambda\langle 1, 1\rangle$, so $x = y = 2$ and $f = 8$. Along the line, $x^2 + y^2$ grows without bound, so this is a minimum. **3.** Minimize the squared distance $x^2 + y^2$. Then $2x = 2\lambda$ and $2y = \lambda$, so $x = 2y$. The constraint gives $4y + y = 5$, so $y = 1$ and $x = 2$. The distance is $\sqrt{4 + 1} = \sqrt{5}$. **4.** $\langle 1, 1\rangle = \lambda\langle 2x, 2y\rangle$, so $x = y$. Then $2x^2 = 2$, so $x = \pm 1$. Compare: $f(1, 1) = 2$ and $f(-1, -1) = -2$. **5.** $y = 2\lambda x$ and $x = 2\lambda y$. Multiply: $xy = 4\lambda^2 xy$. If $xy \neq 0$, then $\lambda = \pm\frac{1}{2}$ and $x^2 = y^2$. The constraint gives $x^2 = 4$. The points $(2, 2)$ and $(-2, -2)$ give $4$; $(2, -2)$ and $(-2, 2)$ give $-4$. If $xy = 0$ then $f = 0$, which is neither extreme. **6.** $2xy = 2\lambda x$ and $x^2 = 2\lambda y$. If $x \neq 0$, the first gives $\lambda = y$, and the second gives $x^2 = 2y^2$. The constraint becomes $3y^2 = 3$, so $y = \pm 1$ and $x = \pm\sqrt{2}$. Then $f = x^2y = 2y$, which is $2$ or $-2$. The case $x = 0$ gives $f = 0$. **7.** $2 = 2\lambda x$ and $3 = 2\lambda y$, so $\frac{y}{x} = \frac{3}{2}$ and $y = \frac{3}{2}x$. The constraint gives $x^2 + \frac{9}{4}x^2 = 13$, so $x^2 = 4$ and $x = \pm 2$. The points $(2, 3)$ and $(-2, -3)$ give $13$ and $-13$. **8.** $yz = \lambda$, $xz = \lambda$ and $xy = \lambda$. Comparing them gives $x = y = z$, so each is $4$ and $f = 64$. **9.** Minimize $x^2 + y^2 + z^2$. Then $\langle 2x, 2y, 2z\rangle = \lambda\langle 1, 2, 2\rangle$, so $(x, y, z) = t(1, 2, 2)$. The constraint gives $t + 4t + 4t = 9$, so $t = 1$. The point is $(1, 2, 2)$ and the distance is $\sqrt{1 + 4 + 4} = 3$. **10.** Minimize $S = 2(xy + yz + xz)$ subject to $xyz = 8$. The equations $2(y + z) = \lambda yz$, $2(x + z) = \lambda xz$ and $2(x + y) = \lambda xy$ give $x = y = z$. Then $x^3 = 8$, so the box is a $2\times 2\times 2$ cube with $S = 24$. **11.** Maximize $xy$ subject to $2x + 2y = 40$. Then $y = 2\lambda$ and $x = 2\lambda$, so $x = y = 10$ and the area is $100$. **12.** In problem 1, $\lambda = x = 5$. Raising the constraint from $10$ to $11$ should raise the max by about $5$, to $30$. The exact max for $x + y = 11$ is $5.5^2 = 30.25$. Close. ## Common mistakes on these problems - **Dividing by a variable that could be $0$.** In problems 5 and 6, check the cases $x = 0$ or $y = 0$ separately. - **Forgetting the constraint equation.** $\nabla f = \lambda\nabla g$ alone has too many unknowns. The constraint is the extra equation. - **Stopping at one point.** On a circle there is both a max and a min. Find every solution and compare. - **Minimizing distance with a square root.** Minimize the squared distance instead. It has the same minimum point and simpler derivatives. ## Keep practicing For unlimited problems with instant feedback, try the [Lagrange multipliers lesson](/learn/calculus-3/partial-derivatives/lagrange-multipliers) and the [critical points and second derivative test lesson](/learn/calculus-3/partial-derivatives/critical-points-and-second-derivative-test). For practice with the gradients these problems need, see the [partial derivative problems](/practice-problems/partial-derivatives). ## Frequently asked questions ### How do you solve the Lagrange equations? Write each component of $\nabla f = \lambda\nabla g$, solve each one for $\lambda$, and set the results equal. That gives a relation like $x = y$ or $x^2 = 2y^2$. Substitute it into the constraint. Watch for cases where a variable is $0$, since dividing by it would lose solutions. ### How do you know if a point is a max or a min? On a closed, bounded constraint like a circle, plug all candidates into $f$ and compare. On an unbounded constraint like a line, think about what happens far away: in problem 2, $x^2 + y^2$ grows without bound along the line, so the single candidate is a minimum. ### What does lambda mean? It is the rate at which the optimal value changes as the constraint value $k$ changes. If $\lambda = 5$, raising $k$ by $1$ raises the best value of $f$ by about $5$. Problem 12 checks this.