Methods for two problems?

[Modigliani]

The first problem involves finding the cost of the cheapest rectangular housing of a machine. I'm given the volume, and the relationship base = 2*width. The top is reinforced and costs more per square unit to make than the sides (bottom is open). I'm not sure what to solve for. Does the pricing have any bearing on what needs to be minimized?

The second involves a cone shaped vat. I'm given height, radius, and the rate at which the liquid is being pumped into it (dv/dt). I have to find how fast the liquid rises at a certain liquid level, and dr/dt at that time as well.

I'm trying to figure these out on my own (why I didn't give numbers), but I don't know what I'm doing, honestly.

 

[Deleted member 4993]

The first problem involves finding the cost of the cheapest rectangular housing of a machine. I'm given the volume, and the relationship base = 2*width. The top is reinforced and costs more per square unit to make than the sides (bottom is open). I'm not sure what to solve for. Does the pricing have any bearing on what needs to be minimized?

Cost of housing involves AREA of the enclosure. However, the relationship given does not make sense (base = 2*width --- base what?).

Assume

base width = W

base length = L

Height = H

Volume = V = W*L*H = constant

Then Area of sides = 2*H*(L+W) = 2*V*(L+W)/(L*W)

Area of the top = L*W

Then calculate cost - apply relationship and continue.....

The second involves a cone shaped vat. I'm given height, radius, and the rate at which the liquid is being pumped into it (dv/dt). I have to find how fast the liquid rises at a certain liquid level, and dr/dt at that time as well.

Draw a picture.

V = (pi)/3 * h *r^2

From the given total volume, height and radius - you can find their respective ratios and then differentiate.

Look up the problem solved at:

viewtopic.php?f=3&t=31884&p=123464&hilit=cone#p123464



I'm trying to figure these out on my own (why I didn't give numbers), but I don't know what I'm doing, honestly.

 

[Modigliani]

"base" meant base length.

 

[mmm4444bot]

… Does the pricing have any bearing on what needs to be minimized? …

Yes. The dimensions of the least-expensive housing will change if the price to manufacture the side and/or top materials changes.

You want to begin by expressing all of the material areas in terms of one dimension (eg: the width).

Write a cost function to manufacture the materials for one housing: C(w).

Find the derivative of the cost function: C '(w).

Find w for C '(w) = 0.

This gives you the width of the least-expensive housing. Use this value of w to find the other two dimensions.

The actual cost for the housing is C(w).

EG:

The volume of an open-bottomed machine housing is four cubic feet. The housing is twice as wide as it is deep.

The side materials cost 16 cents per square foot, and the top material costs $1.13 per square foot.

Find the dimensions (to four decimal places) of the housing that minimizes the cost of materials.

What is the cost for this particular housing's materials?

w = width

2w = depth

h = height

V = l * w * h

4 = 2w * w * h

h = 2/(w^2)

Total area of the four sides = 12/w

Total area of the top = 2w^2

C(w) = 2.26w^2 + 1.92/w

C '(w) = 4.52w - 1.92/w^2

C '(w) = 0

w^3 = 1.92/4.52

w = 0.7517

2w = 1.5034

2/w^2 = 3.5395

C(0.7517) = 3.83

The dimensions of the least-expensive housing are 0.7517 feet wide by 1.5034 feet deep by 3.5395 high.

The materials for this housing cost $3.83.

 

[galactus]

The second involves a cone shaped vat. I'm given height, radius, and the rate at which the liquid is being pumped into it (dv/dt). I have to find how fast the liquid rises at a certain liquid level, and dr/dt at that time as well.

If you need dh/dt, the change in height, then create similar triangles and solve for r in terms of h.

If you want dr/dt, then do the same and solve for h in terms of r.

Let's say we have a cone with height 10 and radius 5. Then by similar triangles, we would have \(\displaystyle \frac{5}{10}=\frac{r}{h}\)

To find dh/dt, solve for r in terms of h and sub into the cone volume formula.

As per our example, we get \(\displaystyle r=\frac{h}{2}\)

Sub into formula: \(\displaystyle V=\frac{\pi}{3}(\frac{h}{2})^{2}h\)

Now, differentiate: \(\displaystyle \frac{dV}{dt}=\frac{\pi}{4}h^{2}\frac{dh}{dt}\)

Plug in your knowns and solve for dh/dt.

To find dr/dt, sub in h=2r.

\(\displaystyle V=\frac{\pi}{3}r^{2}(2r)\)

\(\displaystyle \frac{dV}{dt}=2{\pi}r^{2}\frac{dr}{dt}\)

Now, plug in the knowns and solve for dr/dt.

See?. nothing to it.

Take note, these problems are done to death. What I mean is they are very cliche and used over and over again. If you google or search the sites, you will more than likely find something similar if not identical.