- 11 Mar '08 16:33The three princes of Serendip

Went on a little trip.

They could not carry too much weight;

More than 300 pounds made them hesitate.

They planned to the ounce. When they returned to Ceylon

They discovered that their supplies were just about gone

When, what to their joy, Prince William found

A pile of coconuts on the ground.

"Each will bring 60 rupees," said Prince Richard with a grin

As he almost tripped over a lion skin.

"Look out!" cried Prince Rupert with glee

As he spied some more lion skins under a tree.

"These are worth even more - 300 rupees each

If we can just carry them down to the beach."

Each skin weighed fifteen pounds and each coconut, five,

But they carried them all and made it alive.

The boat back to the island was very small.

15 cubic feet baggage capacity - that was all.

Each lion skin took up one cubic foot

While eight coconuts the same space took.

With everything stowed away they headed to sea

And on the way calculated what their new wealth might be.

"Eureka!" cried Prince Rupert. "Our worth is so great

That there's no other way we could return in this state.

Any other skins or nut which we might have brought

Would now have us poorer. And now I know what -

I'll write my friend Horace in England, for surely

Only he can appreciate our serendipity."

Can you find the number of coconuts and the number of lion skins taken to Serendip? - 11 Mar '08 17:00 / 2 edits

first, 15 = x + 1/8y where x is the number of lion skins and y is the number of coconuts (given by the space requirement. by rearrangement, y = 8(15-x). note: for y to be positive, 0*Originally posted by Swlabr***The three princes of Serendip**

Went on a little trip.

They could not carry too much weight;

More than 300 pounds made them hesitate.

They planned to the ounce. When they returned to Ceylon

They discovered that their supplies were just about gone

When, what to their joy, Prince William found

A pile of coconuts on the ground.

"Each will bring 60 rupees,"

Can you find the number of coconuts and the number of lion skins taken to Serendip? - 11 Mar '08 17:02 / 4 edits

sorry about the repost, i'm not sure why it got cut off....*Originally posted by Aetherael***first, 15 = x + 1/8y where x is the number of lion skins and y is the number of coconuts (given by the space requirement. by rearrangement, y = 8(15-x). note: for y to be positive, 0**

first, 15 = x + 1/8y where x is the number of lion skins and y is the number of coconuts (given by the space requirement.) by rearrangement, y = 8(15-x) = 120-8x. note, for y to be positive, x can vary from 0 to 15.

now we seek to maximize the function: F(x,y) = 300x + 60y, or by substitution:

F(x) = 300x + 60(120-8x) = 300x + 7200 - 480x = 7200 - 180x.

even without calculus, it is obvious that this function, within the domain restriction of x, is clearly maximized at x = 0. Thus, the number of lion skins is 0, and the number of coconuts is 8*15 = 120.

from a more intuitive standpoint, though the lion skins are 5 times as expensive as the coconuts, they take up 8 times as much space. so 8 coconuts will always be a more valuable choice than a single lion skin. kudos on the poem - a nicely setup problem! - 11 Mar '08 17:16So,

pelts are worth 5x more

pelts weight 3x more

pelts occupy 8x more

We want to take as much pelts as we want, but due to occupation restrictions, taking 8 cocunuts is better then 1 pelt. The maximum number of cocunuts it compensates take is maximized by weight restrictions. Optimum number is 12 pelts, 24 coconuts. - 11 Mar '08 17:19

you missed the weight restriction:*Originally posted by Aetherael***sorry about the repost, i'm not sure why it got cut off....**

first, 15 = x + 1/8y where x is the number of lion skins and y is the number of coconuts (given by the space requirement.) by rearrangement, y = 8(15-x) = 120-8x. note, for y to be positive, x can vary from 0 to 15.

now we seek to maximize the function: F(x,y) = 300x + 60y, or by substitu ...[text shortened]... a more valuable choice than a single lion skin. kudos on the poem - a nicely setup problem!

coconuts weight 5, pelts weight 15 and maximum is 300 - 14 Mar '08 09:50 / 2 edits

Number of coconuts x*Originally posted by Swlabr***The three princes of Serendip**

Went on a little trip.

They could not carry too much weight;

More than 300 pounds made them hesitate.

They planned to the ounce. When they returned to Ceylon

They discovered that their supplies were just about gone

When, what to their joy, Prince William found

A pile of coconuts on the ground.

"Each will bring 60 rupees,"

Can you find the number of coconuts and the number of lion skins taken to Serendip?

Numbers of pelts y

I'm assuming they can use the boat only once for free.

Max weight

5x + 15y =< 300

Or

x + 3y =< 60

Profit

60x + 300y = P

Boat restriction

x + 8y =< 120 (removed fractions)

Using both restrictions to eliminate x:

5y =< 60

y =< 12

Using both to eliminate y:

5x =< 120

x =< 24

Max profit at x = 24 and y = 12 giving 60*24 + 300*12 = 5040 rupees. - 15 Mar '08 05:36

That reasoning gives the correct result, but it is wrong.*Originally posted by TheMaster37***Number of coconuts x**

Numbers of pelts y

I'm assuming they can use the boat only once for free.

Max weight

5x + 15y =< 300

Or

x + 3y =< 60

Profit

60x + 300y = P

Boat restriction

x + 8y =< 120 (removed fractions)

Using both restrictions to eliminate x:

5y =< 60

y =< 12

Using both to eliminate y:

5x =< 120

x =< 24

Max profit at x = 24 and y = 12 giving 60*24 + 300*12 = 5040 rupees. - 15 Mar '08 08:32 / 2 edits

Your reasoning is correct. However, if you wanted to maximise 20x-5y, or something like it, where you weren't just looking for the maximum possible values of x and y, your reasoning here would go to pot...*Originally posted by TheMaster37***Number of coconuts x**

Numbers of pelts y

I'm assuming they can use the boat only once for free.

Max weight

5x + 15y =< 300

Or

x + 3y =< 60

Profit

60x + 300y = P

Boat restriction

x + 8y =< 120 (removed fractions)

Using both restrictions to eliminate x:

5y =< 60

y =< 12

Using both to eliminate y:

5x =< 120

x =< 24

Max profit at x = 24 and y = 12 giving 60*24 + 300*12 = 5040 rupees. - 19 Mar '08 17:15

Eight coconuts in a cubic foot?*Originally posted by Swlabr***The three princes of Serendip**

Went on a little trip.

They could not carry too much weight;

More than 300 pounds made them hesitate.

They planned to the ounce. When they returned to Ceylon

They discovered that their supplies were just about gone

When, what to their joy, Prince William found

A pile of coconuts on the ground.

"Each will bring 60 rupees," ...[text shortened]...

Can you find the number of coconuts and the number of lion skins taken to Serendip? - 20 Mar '08 13:25

Not that small.*Originally posted by joe shmo***pretty small coconuts**

Even if each coconut had its own little cell thats still gives it a cube of 6 inches each side (allowing a max length of over 9 inches). With packing bigger coconuts could fit together and quite easily get 8 to a cubic foot.

(Anyone know any Packing Theory?) - 20 Mar '08 14:12As I recall, you use the two inequalities create an area of acceptable combinations, with one variable representing number of coconuts, and the other representing number of lion skins.

Now in order to maximize profit, you need only look at the vertex points of the convex curve formed, and compare the money earned for each.

And with 2 inequalities, you have 3 possibilities.

1) All coconuts. The limiting factor here is weight, as they can only carry 60 coconuts. Selling for 60 each, the three would end up with 3600 in gains.

2) All lion skins. The limiting factor here is the space on the boat. The boat only fits 15 skins. With each selling for 300, that's a gain of 4500.

3) Attempt to maximize both weight and space. With 12 skins and 24 coconuts, they'll carry 300 lbs back and use all 15 cubit feet of space up.

This results in a gain of 300*12 + 60*24 or 5040 total.

Now what if they were to give up a lion skin for more coconuts? In that case, they could only carry 3 more coconuts back, and the net result of the shift would be 3*60 - 300 or a loss of 120 in potential revenue.

So what if they decided to give up coconuts so they could carry an extra lion skin? They would need to give up 8 coconuts to get an extra skin, and the net shift in profit would be 300 - 8*60 or a loss of 180 in revenue.

So 12 lion skins and 24 coconuts is the answer, with a profit of 5040 total. - 20 Mar '08 14:34

This is the right reasoning.*Originally posted by geepamoogle***As I recall, you use the two inequalities create an area of acceptable combinations, with one variable representing number of coconuts, and the other representing number of lion skins.**

Now in order to maximize profit, you need only look at the vertex points of the convex curve formed, and compare the money earned for each.

And with 2 inequalities, ...[text shortened]... 80 in revenue.

So 12 lion skins and 24 coconuts is the answer, with a profit of 5040 total.

Very well explained, thank you!