# Engineering Mathematics Questions and Answers – Lagrange Method of Multiplier to Find Maxima or Minima

This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Lagrange Method of Multiplier to Find Maxima or Minima”.

1. In a simple one-constraint Lagrange multiplier setup, the constraint has to be always one dimension lesser than the objective function.
a) True
b) False

Explanation: This condition is not always necessary because the lesser dimension curve can still be treated as a higher dimension curve.

2. Maximize the function x + y – z = 1 with respect to the constraint xy=36.
a) 0
b) -8
c) 8
d) No Maxima exists

Explanation: Geometrically, we can see that the level curves can go further the origin along the curve xy=36 infinitely and still not reach its maximum value. What the Lagrange multiplier predicts in this case is the minimum value.

3. Which one of these is the right formula for the Lagrange multiplier with more than one constraint?
a) ∇f = (μ)2 * ∇g1 + ∇g2
b) Cannot be applied to more than one constraint function.
c) ∇f = μ * ∇g1 + λ * ∇g2
d) ∇f = μ * ∇g1 + ∇g2

Explanation: The lagrange multiplier can be applied to any number of constraints and the condition is
∇f = Σni=1 μigi
Where μi, μ2 ……..μn are appropriate constraints(scalar multiples).

4. Maximum value of a 3-d plane is to be found over a circular region. Which of the following happens if we increase the radius of the circular region.
a) Maximum value is invariant
b) Maximum value decreases
c) Maximum value increases and minimum value goes lesser
d) minimum value goes higher

Explanation: Consider the level curves of the plane. These are the set of straight lines with equal slope and unequal intercepts. Now as the radius of the circular region is increased, we see that the Lagrange condition(i.e. the level curves to be tangent to the circular boundary) happens to occur further away form the origin. Thus the maximum value is pushed further and the minimum value is decreased further.

5. Find the points on the plane x + y + z = 9 which are closest to origin.
a) (3,3,3)
b) (2,1,3)
c) (2,2,2)
d) (3,4,1)

Explanation: The objective function is f(x,y,z) = x2 + y2 + z2
compute gradient ∇f = 2x i + 2y j + 2z k
Now compute gradient of the function x + y + z = 9
which is
= i + j + k
Using Lagrange condition we have
∇f = λ . ∇g
2x i + 2y j + 2z k = λ * (i + j + k)
⇒ x = y =z
Put this back into constraint function we get
3x = 9 ⇒ (x,y,z) = (3,3,3).
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6. Consider the points closest to the origin on the planes x + y + z = a.
a) The closest point travels farther as a is increased
b) The closest point travels nearer as a is increased
c) The closest point is independent of a as a is not there in the expression of the gradient.
d) Varies as a2, away from the origin.

Explanation: The intercept of the planes increase as we increase the a value. Hence, we may conclude that the closest point of lower a value plane would be closer to the origin. The Lagrange multiplier set up can be used to verified this.

7. The span of a Astroid is increased along both the x and y axes equally. Then the maximum value of: z = x + y along the Astroid is?
a) Increases
b) Decreases
c) Invariant
d) The scaling of Astroid is irrelevant

Explanation: Calculating the gradients considering the general form of Astroid as x2/3 + y2/3 = a2/3 and then equating them by Lagrange condition.
we can conclude that the maximum value increases.

8. The extreme value of the function f(x1, x2,….. xn)=$$\frac{x_1}{2^0}+\frac{x_2}{2^1}+……+\frac{x_n}{2^{n-1}}$$ With respect to the constraint Σmi=1 (xi)2 = 1 where m always stays lesser than n and as m,n tends to infinity is?
a) 1
b) $$\frac{2}{3\sqrt{3}}$$
c) 2
d) 1 ⁄ 2

Explanation: First consider these functions as infinite dimension vectors. Given the constraint dimension is always less than the objective we can apply the Lagrange condition. We now have
$$\frac{1}{1}i_1+\frac{1}{2}i_2+\frac{1}{4}i_3+…..\infty=\lambda.(2x_1i_1+2x_2i_2+2x_3i_3+…\infty)$$
Treating equations individually we get
$$x_1=\frac{1}{2\lambda}$$
$$x_2=\frac{1}{4\lambda}$$
and so on.
Putting back in constraint we get
$$1=\frac{1}{(2\lambda)^2}+\frac{1}{(4\lambda)^2}+….\infty$$
$$1=\frac{1}{(\lambda)^2}+(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{8^2}….\infty)$$
$$\Rightarrow \lambda = \frac{1}{\sqrt{3}}$$
Hence, we get the extreme value (after putting back values of variables in the function) as
extreme value = $$\frac{1}{2\sqrt{3}}+\frac{1}{2\sqrt[3]{3}}+\frac{1}{2\sqrt[5]{3}}+…\infty$$
extreme value = $$\frac{\frac{1}{2\sqrt{3}}}{1-\frac{1}{4}}=\frac{2}{3\sqrt{3}}$$

Sanfoundry Global Education & Learning Series – Engineering Mathematics.

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