Calculus- Optimization Problems

[Rosalina]

Hi,

I'm having trouble determining what equations I should use for the following two problems:

1) The base of a chest is a rectangle, which is twice as long as it is wide. The top, front, and sides are made of oak, and the back and base are made of pine. The chest has a volume of 12.25 dm^3. Oak costs three times as much as pine. Find the dimensions of the box for which the lumber has the lowest cost. (Neglect any effects due to thickness of the sides, top, or base)

2) 10 000 kg of apples turn bad at a rate of 200 kg per month. The September price is 25 cents/kg, and the price increases at 5 cents/kg per month. Storage costs are $350 per month. If the apples are placed in storage in September, when should they be sold for maximum gross return?

I would appreciate any help whatsoever!

[FChaosi_]

1) The first thing to notice is that we can essentially ignore what the cost is, because the cost is is minimised when the total surface area of the box is minimised.
The next thing to notice is that our area function will be of the following form:

A(x) = (1 + 1/3)*A(top) + (1 + 1/3)*A(front) + 2*A(side)

This is simply because the cost of an oak piece is exactly three times the cost of a pine piece with the same area.
Let x be the shortest side of the base (and hence the longer side is 2x). So, A(base) = 2x^2.

We can find the height, h, because:
12.25 = h*A(base)
h = 12.25/(2x^2) = 6.125/x^2

Hence A(x) becomes:
A(x) = 4/3[2x^2] + 4/3[x*6.125/x^2] + 2[2x*6.125/x^2]
Simplifying, A(x) = (2/3)x^2 + [49/(6x)] + [49/(2x)]
A(x) is minimized where A'(x) = 0.
Using a calculator, because differentiating by hand is gonna result in messy fractions, yields a minimum for the area (and so a minumum for the cost) when x = 2.90 dm, to 3 significant figures. So that's the shortest side, the longest side is 2*2.90 = 5.8 dm, and the height is 0.728 dm.

That was a pretty fun problem.

Working on 2).

[kw0573]

An interesting decision in question 1 is deciding whether front face has dimensions h * x or h * 2x. it would seem to me that the cost is minimized if the cheaper wood is used for the greater area, so front (being made from both pine and oak) should be large, at h * 2x, leaving the smaller h * x for the side face that is to be constructed from exclusively the expensive oak.

[TheNintendoChip]

I'm going to give my input on this, and I too agree that #1 is a little tricky.

First, @kw0573 is correct in that there are two ways to tackle this problem. The first is to consider the sides as having dimensions h by 2x and the front & back as having dimensions h by x. The second way is to just reverse the two dimensions. @FChaosi_ has done the method where the sides have dimensions h by 2x, although...

5 hours ago, FChaosi_ said:

A(x) = 4/3[2x^2] + 4/3[x*6.125/x^2] + 2[2x*6.125/x^2]
Simplifying, A(x) = (2/3)x^2 + [49/(6x)] + [49/(2x)]
 

This step is incorrect. The leading term should be 8/3x^2, not 2/3! Finding A'(x), setting A'(x)=0, and solving for x yields x=(49)^(1/3)/2, and h happens to be equal to this. The total cost function I wrote is: C(x)= x(2x)(3c)+x(h)(3c)+2x(h)(3c)(2)+x(h)(c)+x(2x)(c). This is equivalent to what @FChaosi_ has for the surface area multiplied by 3c, where c is the cost of the pine wood. I also multiplied by 3 to remove the fractions. Let c=0.6 (just a random test number) and find C(49^(1/3)/2)=48.2058658059.

The method I used is to where the sides have dimensions h by x, which is also the method that @kw0573 seems to think is correct. This is implied (or was at least to me) as the "long" side would be used for the front (silly technicality, I know, but this is IB we're talking about). The cost function for this is similar to the one above and is given by C(x)= x(2x)(3c)+2x(h)(3c)+x(h)(3c)(2)+2x(h)(c)+x(2x)(c). We can make the same substitution for h, which is 12.25/(2x^2), and simplify C(x) to be: (12.25/x)(7c)+8(x^2)(c). C'(x)=c(16x-49(7)/(4x^2)). Solving for x gives 7/4, and h is 2. Much nicer numbers than the first! And if we let c=0.6 like before, and do C(7/4), we find that it is 44.1. This is lower than the first cost, so we should accept the dimensions x=7/4 dm and h=2 dm. 

 

For #2, we're going to write a bunch of smaller functions to model what's going on. First, let's write one for the total amount of apples after x months have passed (since September). This will be A(x)=10000-200x, as the amount decreases linearly, and 200 decay every month. Next let's write the price that we can sell them at after x months. In September, this is just 25, but it increases (linearly, again) by 5 cents per month. This gives P(x)=25+5x. Now let's look at the storage cost. This is going to be C(x)=350(100)x. I multiplied by 100 to convert to cents, because the cost is in dollars. 

The gross return will then be G(x)=A(x)(P(x))-C(x). In words, this tells us we have A(x) many apples selling for P(x) cents per apple, and we subtract the monthly storage cost. Let's simplify G(x) before we differentiate it. G(x)=(10000-200x)(25+5x)-350x=1000(50-x)(5+x)-35000x=1000(250+45x-x^2)-35000x. G'(x)=1000(45-2x)-35000. Setting G(x)=0 and solving for x, we get x=5 months, so next February.*
*Disclaimer: I was never really that good at these cost problems, so something may be wrong here, but this is what makes sense to me!

Edited January 1, 2017 by TheNintendoChip