anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • 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
  • anonymous
i dun think this is calculus level lol
anonymous
  • anonymous
It could be if it's in an optimization section.

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anonymous
  • anonymous
true
anonymous
  • anonymous
it is an optimization. i have 7 of them to do
anonymous
  • anonymous
Alright - we'll use calculus then. :)
anonymous
  • anonymous
So we've got: \[P=32m-m^4\]
anonymous
  • anonymous
Now, you're ready to take a derivative and set it equal to zero to find your mins and maxes.
anonymous
  • anonymous
When you take the derivative, what do you get?
anonymous
  • anonymous
so it would be 0=32-4m^3
anonymous
  • anonymous
Now, solve for m.
anonymous
  • 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
  • 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
  • anonymous
Any questions, Vince? Did that make sense?
anonymous
  • anonymous
that looks about right
anonymous
  • anonymous
You there Vince?
anonymous
  • anonymous
yeah im going over it real quick
anonymous
  • anonymous
Sounds good! :)
anonymous
  • anonymous
How would you find the minimum?
anonymous
  • anonymous
Do you have a graphing calculator with you?
anonymous
  • anonymous
No. Not allowed to use those in this course for some reason.
anonymous
  • anonymous
OK... we'll take a different appraoch...
anonymous
  • 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
  • anonymous
No =(
anonymous
  • 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
  • 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
  • anonymous
Ya know... to find the minimum... actually, I'm not sure.
anonymous
  • 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
  • anonymous
If it had a minimum, we would have found more than one solution when you set the derivative = 0.
anonymous
  • 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
  • anonymous
Yes, because the initial problem that exists is that we have 2 different variables.
anonymous
  • 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
  • anonymous
sorry for being lengthy i just want to make sure im doing it right
anonymous
  • anonymous
Yeah, that works great! :)
anonymous
  • anonymous
alrighty..so in the problem with x+y^3=32 and P=x(32-x^3)..should the same concept apply
anonymous
  • anonymous
Let's try it and see what happens.
anonymous
  • anonymous
because i get x=32-y^3 and after plugging that it..its get scary lol
anonymous
  • anonymous
right now im at P= 32-(32-y^3)^3)
anonymous
  • anonymous
and im stuck
anonymous
  • anonymous
whoops that should be P=32-y^3(32-(32-y^3)^3)
anonymous
  • 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
  • anonymous
P= 32 - y^3(32-(32-y^3)^3) is what i got original was P=x(32-x)
anonymous
  • anonymous
P=x(32-x^3) i mean
anonymous
  • anonymous
Where is the (32-x^3) coming from?
anonymous
  • 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
  • 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
  • anonymous
im not solving for y im solving for x which is x=32-y^3
anonymous
  • anonymous
Ok.
anonymous
  • anonymous
so i can use x in the formula of P=x(32-x^3)
anonymous
  • anonymous
see what i mean
anonymous
  • anonymous
I do.
anonymous
  • anonymous
What I'm having trouble seeing, is where you go from there...
anonymous
  • anonymous
Let me see if I understand this...
anonymous
  • anonymous
wait isnt minimun a positive number and maximum a negative number?
anonymous
  • anonymous
It depends, really.
anonymous
  • anonymous
Minimum should be a "smallest possible" product. Maximum should be a "largest possible" product.
anonymous
  • anonymous
\[P=x(32-x^3)\] \[P=(32-y^3)(32-(32-y^3)^3)\]
anonymous
  • anonymous
yeah..now thats where im stuck i dont know what to do with that
anonymous
  • anonymous
could tou take the derivative of it?
anonymous
  • 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
  • anonymous
Then, take the derivative of :\[P=1024-32y^3-(32-y^3)^4\]Not terribly nice, but better than before...
anonymous
  • anonymous
yeah but then how do we solve that when it equals to 0?
anonymous
  • anonymous
Let's cross that bridge after we take the derivative
anonymous
  • anonymous
what did you get?
anonymous
  • 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
  • anonymous
\[P=-96y^2+12y^2(32-y^3)^3\]
anonymous
  • 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
  • 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
  • anonymous
I really don't think that's right though. I like the first way it was done.
anonymous
  • 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
  • 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
  • anonymous
lol well we will just go with that now i must venture into finding the min

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