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.

- anonymous

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- anonymous

\[n+m^3=32\]
\[P=nm\]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. :)

- anonymous

i dun think this is calculus level lol

- anonymous

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

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## More answers

- anonymous

true

- anonymous

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

- anonymous

Alright - we'll use calculus then. :)

- anonymous

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

- anonymous

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

- anonymous

When you take the derivative, what do you get?

- anonymous

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

- anonymous

Now, solve for m.

- anonymous

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

- anonymous

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.

- anonymous

Any questions, Vince? Did that make sense?

- anonymous

that looks about right

- anonymous

You there Vince?

- anonymous

yeah im going over it real quick

- anonymous

Sounds good! :)

- anonymous

How would you find the minimum?

- anonymous

Do you have a graphing calculator with you?

- anonymous

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

- anonymous

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

- anonymous

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?

- anonymous

No =(

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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?

- anonymous

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

- anonymous

Yeah, that works great! :)

- anonymous

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

- anonymous

Let's try it and see what happens.

- anonymous

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

- anonymous

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

- anonymous

and im stuck

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Ok.

- anonymous

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

- anonymous

see what i mean

- anonymous

I do.

- anonymous

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

- anonymous

Let me see if I understand this...

- anonymous

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

- anonymous

It depends, really.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

could tou take the derivative of it?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Let's cross that bridge after we take the derivative

- anonymous

what did you get?

- anonymous

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

- anonymous

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

- anonymous

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 \[0=-8+(32-y^3)^3\]

- anonymous

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:
\[\sqrt[3]{30}=\sqrt[3]{y^3}\]
\[\sqrt[3]{30}=y\]

- anonymous

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

- anonymous

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.

- anonymous

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!! :)

- anonymous

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

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