Question

1. local max, local min, and saddle point of
f (x, y) = x^3 + 8y^3 - 6xy
2. A rectangular box, without a lid made 12m^2 of cardboard. Find the max volume of such a box?

Answer 1

Partial differentiation wrt x
\[\delta f / \delta x = 6x ^{2}-6y \]

Equating this to 0 we get => x^2 = y

Partial differentiation wrt y
\[\delta f / \delta y = 48x ^{2}-6y \]

Equating this to 0 we get => 4y^2 = x

Solving this two equations we get (0,0), (0.63, 0.4) and (0.63, -0.4)

Now \[\delta ^{2} f / \delta x ^{2} = 12x\]

And
\[\delta ^{2} f / \delta y ^{2} = 48y\]

And \[\delta ^{2} f / \delta y \delta x = -6\]

To discover the nature of the turning point @ point (x,y), the nature of the roots of the following equations are to be checked for lamda
\[(\delta ^{2} f / \delta x ^{2} - \lambda)(\delta ^{2} f / \delta y ^{2} - \lambda) - (\delta ^{2} f / \delta y \delta x)^2 = 0\] -

at (0,0)

\[(-\lambda)((-\lambda) - (-6)^{2} = 0 => \lambda = \pm6\]

Since, the roots are of opposite sign hence it is a saddle point

at (0.63,0.4)

\[(12x-\lambda)((48y-\lambda) - (-6)^{2} = 0 \]

Both of these roots are positive, hence there is local minimum,

Similarly for ((0.63,-0.4), both roots are negative, hence there is a local maximum

Answer 2

For question 2, let the three sides of the box be x, y and z. Hence the volume is f(x,y,z) = xyz
The constraint function
The cupboard side is 12m^2. Now for all side the total area of the walls is
g(x,y,z) = (2xy + 2yz + 2zx - 12)

Hence if the Lagrangian function be L and multiplier be lambda, then

\[L = f(x,y,z) - \lambda g(x,y,z)\]

The partial differentiations are to be equated to zeroes -

\[\delta L / \delta x = yz - 2\lambda (y+z) = 0 => \lambda = yz / 2(y+z)\]
Similarly,
\[\delta L / \delta y = zx - 2\lambda (z+x) = 0 => \lambda = zx / 2(z+x)\]
\[\delta L / \delta z = xy - 2\lambda (x+y) = 0 => \lambda = xy / 2(x+y)\]

Solving all these reaches => x=y=z

Again, \[\delta L / \delta \lambda = 2xy + 2yz + 2zx - 12 = 0 => 6x^2 = 12 => x = \sqrt{2}\]

Hence\[V = (\sqrt{2})^{3} = 2\sqrt{2} \approx 2.83m\]

Answer 3

sorry the unit should be m^3

Answer 4

Please let me know if you need any explanation of any step

Answer 5

owh i'll need time to read up 1st. because i was thrown with this questions by my lecturer when he haven even started the calculus topic. anyway thank you very much. =)

Answer 6

actually these are little advanced topics, calculus generally should start with idea of limit

Answer 7

erm, owh then i should rephrase it as, he started the basics of calculus haha. still at early stages. probably he wanted us to do some self study.

Answer 8

yeah, cheers...

Answer 9

just one thing can you see the posted equation in the correct form

Answer 10

yeah they are very neat. thank you very much for spending your time on it.

Answer 11

welcome, i think there is some problem with my machine though :)

Answer 12

erm, based on © 2007 Paul Dawkins' notes, Q1 should give (0,0) as saddle, and (1,1/2) as minimum, while Q2 gives volume of \[3/2\sqrt{6}\]

Answer 13

I agree with the saddle point before, the note you are referring is correct on local minima.

(I made a mistake in taking the partial differentiation wrt x, thank you for pointing out)

And I feel the volume is is correct as per my calculation. Can you please give the length of each side of the box as per the note

Answer 14

erm the question is not from the notes, i was just searching the net for some notes on lagrange multiplier, and i found paul's notes and apply it to the question. i think the difference is just at you're using 2xy+2yz+2xz as the g(x), where the question i mentioned is concerning on a box without lid. sorry for not bolding it. lol

Answer 15

Hey, you are correct, I missed it. again!!! I jumped quite a few guns here :)