Question

hope someone can help me. The sum of a number and the cube of a second number is 32. The product of the two number, x and 32-x^3, is P= x(32-x^3). Find the maximum product.

Answer 1

\[n+m^3=32\]
\[P=nm\]Solve the first equation for n and plug it into the second equation.
\[P=m(32-m^3)\]Now, to maximize this, we can do it one of two ways - depending on whether or not you're in calculus. :)

Answer 2

i dun think this is calculus level lol

Answer 3

It could be if it's in an optimization section.

Answer 4

true

Answer 5

it is an optimization. i have 7 of them to do

Answer 6

Alright - we'll use calculus then. :)

Answer 7

So we've got:
\[P=32m-m^4\]

Answer 8

Now, you're ready to take a derivative and set it equal to zero to find your mins and maxes.

Answer 9

When you take the derivative, what do you get?

Answer 10

so it would be 0=32-4m^3

Answer 11

Now, solve for m.

Answer 12

\[32-4m^3=0\]Subtract 32 on both sides
\[-4m^3=-32\]Divide -4 on both sides
\[m^3=8\]Cube root both sides
\[m=2\]

Answer 13

To find the max product, however... you have to take the 2 we found and plug it back in the first equation for m.

\[n+m^3=32\]Well, now we know that m = 2
\[n+(2)^3=32\]Cube it
\[n+8=32\]Solve for n
\[n+8=32\]Subtract 8 on both sides
\[n=24\]

Now that we know m = 8 and n = 24, we can determine their product.
(8)(24)=192

So 192 is the max product.

Answer 14

Any questions, Vince? Did that make sense?

Answer 15

that looks about right

Answer 16

You there Vince?

Answer 17

yeah im going over it real quick

Answer 18

Sounds good! :)

Answer 19

How would you find the minimum?

Answer 20

Do you have a graphing calculator with you?

Answer 21

No. Not allowed to use those in this course for some reason.

Answer 22

OK... we'll take a different appraoch...

Answer 23

Let's check out where this function has zeros.
\[32-4x^3=0\]Divide everything by 4.
\[8-x^3=0\]Now, what we're looking at is the difference of two cubes. Do you remember how to factor those?

Answer 24

No =(

Answer 25

That's ok - not many do.
\[2^3-x^3=0\]Factors as:
\[(2-x)(2^2+2x+x^2)=0\]Which becomes:
\[2-x=0\]\[x^2+2x+4=0\]

Answer 26

If you solve the top equation, you get x=2
If you solve the bottom equation you end up with complex solutions
So this function only has 1 real zero; at x=2

Answer 27

Ya know... to find the minimum... actually, I'm not sure.

Answer 28

I was hoping that factoring it would help us out, by showing us that this thing has a solution with a multiplicity higher than 1at that maximum we found, but it didn't.

Answer 29

If it had a minimum, we would have found more than one solution when you set the derivative = 0.

Answer 30

Hmm.. another thing i was looking at is when we solved for the maximum..i have mine set up as x+y^3=32. are you supposed to solve for x in that equation and then use that in the P=x(32-x^3) ? thats what the teacher showed in class with another problem with sum and product.

Answer 31

Yes, because the initial problem that exists is that we have 2 different variables.

Answer 32

okay.. well in the previous problem it was. If the sum of two numbers is 32, the product of the two numbers, x and 32-x, is P=x(32-x). Find the maximum product. I solved for x in x+y=32, which was x= 32-y. Then plugged in P=32-y(32-32-y), which i got P=-32y + y^2. Then i found the derivative and got -32+2y=0. Solving it for y=16. Plugging the 16 i got into x+y=32, which is x+16=32, x=16. Then i plugged the x value in P=x(32-x) and got P=256. Is that right or did i do something wrong?

Answer 33

sorry for being lengthy i just want to make sure im doing it right

Answer 34

Yeah, that works great! :)

Answer 35

alrighty..so in the problem with x+y^3=32 and P=x(32-x^3)..should the same concept apply

Answer 36

Let's try it and see what happens.

Answer 37

because i get x=32-y^3 and after plugging that it..its get scary lol

Answer 38

right now im at P= 32-(32-y^3)^3)

Answer 39

and im stuck

Answer 40

whoops that should be P=32-y^3(32-(32-y^3)^3)

Answer 41

\[x+y^3=32\]\[x=32-y^3\]And tell me if this looks right:\[P=32-y^3(32-y^3)^3\]

Answer 42

P= 32 - y^3(32-(32-y^3)^3) is what i got
original was P=x(32-x)

Answer 43

P=x(32-x^3) i mean

Answer 44

Where is the (32-x^3) coming from?

Answer 45

original formula is P=x(32-x^3)...im just plugging in x which is 32-y^3 from what we found in x + y^3=32

Answer 46

If we solved the original equation for y, we should get:\[x+y^3=32\]\[y^3=32-x\]\[\sqrt[3]{y^3}=\sqrt[3]{32-x}\]Leaving us with:\[y=\sqrt[3]{32-x}\]Not y=32-x^3

Answer 47

im not solving for y im solving for x which is x=32-y^3

Answer 48

Ok.

Answer 49

so i can use x in the formula of P=x(32-x^3)

Answer 50

see what i mean

Answer 51

I do.

Answer 52

What I'm having trouble seeing, is where you go from there...

Answer 53

Let me see if I understand this...

Answer 54

wait isnt minimun a positive number and maximum a negative number?

Answer 55

It depends, really.

Answer 56

Minimum should be a "smallest possible" product. Maximum should be a "largest possible" product.

Answer 57

\[P=x(32-x^3)\]
\[P=(32-y^3)(32-(32-y^3)^3)\]

Answer 58

yeah..now thats where im stuck
i dont know what to do with that

Answer 59

could tou take the derivative of it?

Answer 60

Sweet Jesus that is ugly...

What do you say we do this instead:
\[P=x(32-x^3)\]\[P=32x-x^4\]Then Substitute:
\[P=32(32-y^3)-(32-y^3)^4\]
\[P=1024-32y^3-(32-y^3)^4\]

Answer 61

Then, take the derivative of :\[P=1024-32y^3-(32-y^3)^4\]Not terribly nice, but better than before...

Answer 62

yeah but then how do we solve that when it equals to 0?

Answer 63

Let's cross that bridge after we take the derivative

Answer 64

what did you get?

Answer 65

\[P=1024-32y^3-(32-y^3)^4\]
\[P=-32(3)y^2-(4)(32-y^3)^3(-3y^2)\]And it will need to be cleaned up some...

Answer 66

\[P=-96y^2+12y^2(32-y^3)^3\]

Answer 67

Set equal to zero:\[0=-96y^2+12y^2(32-y^3)^3\]We can factor:\[0=12y^2(-8+(32-y^3)^3)\]Then use the Zero-Product Property:\[0=12y^2\]and \[0=-8+(32-y^3)^3\]

Answer 68

So the top one gives us 2 zeros.

\[0=-8+(32-y^3)^3\]Here, add 8 to both sides:
\[8=(32-y^3)^3\]Cube root both sides:
\[2=32-y^3\]Subtract 32 on both sides:
\[-30=-y^3\]Divide -1 on both sides:
\[30=y^3\]Cube root both sides:
\[\sqrt[3]{30}=\sqrt[3]{y^3}\]
\[\sqrt[3]{30}=y\]

Answer 69

I really don't think that's right though. I like the first way it was done.

Answer 70

Oops... I found a mistake I made the first time way above. I said m = 8 and n = 24.

m isn't 8, m^3 is 8. m is really 2.

So the max product should be (2)(24)=48.

Answer 71

And with that, I have got to head to bed. We gave his one a valliant attempt. Check with your teacher. See if you can get him to explain his method again. Good luck to you, as well!! :)

Answer 72

lol well we will just go with that now i must venture into finding the min