Question

A channel is to be constructed out of a sheet of metal, to allow gravel to be removed from a mining site. The sheet of metal has a width 40cm, and each side is to be bent up (at the same angle theta=pi/4 from the horizontal) so as to maximise the amount of gravel that the channel can carry

Answer 1

Have been working on it all day still got no-where.

Answer 2

Can you write an expression for the cross-sectional area as a function of \(w\)?

Answer 3

Have to find the portion -w- on the sheet metal as well as calculate maximum cross sectional area of the channel.

Answer 4

The length of the top side is: (2wcos(45) + L) = 2 * w / sqrt(2) + L
Area of trapezium A = 1/2 * (L + 2*w/sqrt(2) + L) * w / sqrt(2)
A = 1/2 * (2L + 2*w/sqrt(2)) * w/sqrt(2) = (L + w/sqrt(2)) * w /sqrt(2)
A = Lw/sqrt(2) + w^2/2
Put L = 40 - 2w in A.
A = w/sqrt(2) * (40 - 2w) + w^2/2

THIS is as far as ive gotten.

Answer 5

Okay, I agree with that expression of the area. Do you recognize the form of the equation?

Answer 6

It's a parabola. Find the vertex and you've got your solution.

Vertex of a parabola in the form \(ax^2+bx+c\) is at \(x = -\dfrac{b}{2a}\)

Answer 7

The vertex gives me my maximum cross sectional area correct?

Answer 8

with the equation written as you have it, the vertex of the parabola is at the value of \(w\) that gives the maximum cross-sectional area

Answer 9

Slightly confused. How do i isolate the two?

Answer 10

isolate which two?

Answer 11

Ive been told to continue with that, find the derivative and make it equate to 0 to get max. Then solve for w.

Answer 12

well, okay, if you want to do it the longer way, but for any parabola of the form \(y = ax^2+bx+c\), the vertex is at \(x = -\frac{b}{2a}\) which should be able to convince yourself of by taking the derivative of the parabola equation, setting it equal to 0 and solving for \(x\).

Answer 13

Can you show me how id go by doing that? still confused. sorry.

Answer 14

You have \[A = \frac{1}{2}w^2-\sqrt{2}w^2 + 20\sqrt{2} w = (\frac{1}{2}-\sqrt{2})w^2 + 20\sqrt{2}w\]\[a = (\frac{1}{2}-\sqrt{2})\]\[b = 20\sqrt{2}\]vertex is located at \[w=-\frac{20\sqrt{2}}{2(\frac{1}{2}-\sqrt{2})}\]sharpen your pencil and simplify that that...

Answer 15

Giving w = 15.469 (3dp). Do i take this sub it back into my original formula of A = And allgood?

Answer 16

yes, that would give you the cross-sectional area

Answer 17

1407.827 sound correct?

Answer 18

Here's a plot of the area function as w goes from 0 to 31.

No, 1408 does not sound correct. If we had a rectangular channel of width 40, it would have to be approximately 35 tall to have that cross-sectional area!

Answer 19

106.857(3dp)?

Answer 20

no. how are you evaluating this?

Answer 21

Take \(w= 15.469\). Square it. Multiply it by \(a = \frac{1}{2}-\sqrt{2} \approx -0.914\)
Take \(w = 15.469\). Multiply it by \(b = 20\sqrt{2} \approx 28.284\)
Add the two together. What do you get?

Answer 22

compare your answer with the y-value of the vertex of the parabola in the graph I gave you. the vertical line indicates the position of the vertex, if it isn't clear

Answer 23

it's like 4AM here. time to hit the sack. good luck wrapping up the problem!

Answer 24

209.012

Answer 25

^ WOW ahaha.

Answer 26

still off by about 10, I'm not sure what you're doing wrong

Answer 27

No worries. Thanks for your help.