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# Partial differential equations/directional derivatives - rep given for help! watch

1. If you can do one or more than one of these problems and show your working, I will be grateful and rep you. I am currently trying to revise for an exam in differential equations and am having troubles revising this area.

1. The material used to make the bottom of a rectangular box is twice as expensive (per unit of its area) as that used to make the top or sides. What are the dimensions of the cheapest box of 12cm^3?

2. Find the rate of change of the function in the direction of the vector 'v' given:

a) f(x,y) = x/(1+y) for v = i - j

b) f(x,y) = x^2 + y^2 for v = (1, -2) making a positive 60 degree angle with the x-axis.
2. If you cant actually help, then pointing out useful sites for me to research would also help!
3. (Original post by RobbieC)
1. The material used to make the bottom of a rectangular box is twice as expensive (per unit of its area) as that used to make the top or sides. What are the dimensions of the cheapest box of 12cm^3?
Call the height, width and length of the box H, W, L.

Then we know HWL = 12. We'll use this to eliminate L with L = 12/(HW)

Also the cost of the box (assuming it has a top) is proportional to

C = 2WL + WL + HL + HL + HW + HW
= 3WL + 2HL + 2HW
= 36/H + 24/W + 2HW.

Work out dC/dH and dC/dW (partial derivatives). At the minimum these are both zero.

2. Find the rate of change of the function in the direction of the vector 'v' given:

a) f(x,y) = x/(1+y) for v = i - j

b) f(x,y) = x^2 + y^2 for v = (1, -2) making a positive 60 degree angle with the x-axis.
The rate of change of a function f(x,y) at a point p in the direction v is

. denotes the dot product
|v| is the modulus of v
4. Thanks very much for the help. A late nighter allowed me to solve 2, though im not sure how right it is! But that was a big help on 1. Seems so obvious thinking about it now.

I guess hindsight is 20x20, though.

I will rep you when my power comes back neo!

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Updated: February 20, 2006
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