Solve problems with limits or ?

[prehisto]

Homework Statement

I have a series of problems
1.problem. When this number is summed with its square. The summ is the least possible.
2.problem Find positive number which substraction with its third power would be the greatest.
3.problem Find to numbers which x+y=5 but multiplication is the least

Homework Equations

1)x+x^2=
2)x-x^3=
3)x+y=5, x*y=

The Attempt at a Solution

I suppose I should you use limits but can't figure out how ( what kind of limit i should use) Some-kind of help ,please?

 

[Mastermind01]

Do you know/ can you use derivatives?

 

[Mark44]

Homework Statement

I have a series of problems
1.problem. When this number is summed with its square. The summ is the least possible.
2.problem Find positive number which substraction with its third power would be the greatest.
3.problem Find to numbers which x+y=5 but multiplication is the least

Homework Equations

1)x+x^2=
2)x-x^3=
3)x+y=5, x*y=

The Attempt at a Solution

I suppose I should you use limits but can't figure out how ( what kind of limit i should use) Some-kind of help ,please?

These problems have nothing to do with limits. Problems 1 and 3 can be done without calculus. Problem 2 can be done using calculus, or you can find an approximate solution by graphing an equation.

You posted this in the Precalc section, so our assumption is that you are in a math class that isn't at the calculus level. What class are you currently in?

 

[prehisto]

Do you know/ can you use derivatives?

Yes,I can

 

[prehisto]

These problems have nothing to do with limits. Problems 1 and 3 can be done without calculus. Problem 2 can be done using calculus, or you can find an approximate solution by graphing an equation.

You posted this in the Precalc section, so our assumption is that you are in a math class that isn't at the calculus level. What class are you currently in?

I rely do not understand the distinction between precalc an calculus. I just assumed that this is not "advanced" mathematics.

I know derivatives, integration and differential solutions. I just don't understand how to solve this kind of problems.

 

[Mark44]

I know derivatives, integration and differential solutions.

Precalc includes algebra and trigonometry - math areas that are usually taught before ("pre-") calculus.

Do you know how to use derivatives to find the minimum or maximum value of a function?

I am moving this thread to the Calculus & Beyond section.

 

[Mastermind01]

Yes,I can

Like Mark44 said this is then a simple maxima minima problem.

 

[prehisto]

Precalc includes algebra and trigonometry - math areas that are usually taught before ("pre-") calculus.

Do you know how to use derivatives to find the minimum or maximum value of a function?

I am moving this thread to the Calculus & Beyond section.

Yes, thank you for hint and already solved these problems,but i have another one which is a bit confusing.

I have to find 2 positive numbers x+y=10 in such way x^2/2 +y^3=0 is the least.
I plotted both functions and they have a cross point only when y is negative (15;-5). Should i plot x+y=10 and derivative function of x^2/2 +y^3=0 ?

 

[Dick]

Yes, thank you for hint and already solved these problems,but i have another one which is a bit confusing.

I have to find 2 positive numbers x+y=10 in such way x^2/2 +y^3=0 is the least.
I plotted both functions and they have a cross point only when y is negative (15;-5). Should i plot x+y=10 and derivative function of x^2/2 +y^3=0 ?

I'm assuming the problem is to minimize $x^2/2+y^3$ (no $=0$, that wouldn't make much sense). Since you know $x+y=10$ then you also know $y=10-x$. Substitute that into the quadratic to get a function of the single variable $x$.

 

[prehisto]

I'm assuming the problem is to minimize $x^2/2+y^3$ (no $=0$, that wouldn't make much sense). Since you know $x+y=10$ then you also know $y=10-x$. Substitute that into the quadratic to get a function of the single variable $x$.

If i do this, I end up getting x =14,7 that means y= -4,7 which does not comply with the condition that both numbers should be positive.

 

[Dick]

If i do this, I end up getting x =14,7 that means y= -4,7 which does not comply with the condition that both numbers should be positive.

I don't think that is correct. Can you show your work?

 

[Mark44]

When you solve a quadratic equation, there will be two solutions. In this case, neither of the solutions is x = 14.7, so I concur with Dick's statement.

 

[prehisto]

When you solve a quadratic equation, there will be two solutions. In this case, neither of the solutions is x = 14.7, so I concur with Dick's statement.

You mean x(-x²+61/2 x+700)?
If I solve -x²+61/2 x+700 I also get x values which do not comply with the condition

 

[Mark44]

You mean x(-x²+61/2 x+700)?

Where did this come from?

It would be helpful if you showed us all of your work.

If I solve -x²+61/2 x+700 I also get x values which do not comply with the condition

-x²+61/2 x+700 is not an equation, so you can't solve it. An equation has = between two expressions.

 

[Ray Vickson]

If i do this, I end up getting x =14,7 that means y= -4,7 which does not comply with the condition that both numbers should be positive.

That is not the solution. Start again.

 

[Mark44]

I have to find 2 positive numbers x+y=10 in such way x^2/2 +y^3=0 is the least.
I plotted both functions and they have a cross point only when y is negative (15;-5). Should i plot x+y=10 and derivative function of x^2/2 +y^3=0 ?

I'm pretty sure you have not given the problem correctly. Here's what I believe the problem actually is:

Find two positive numbers x and y, whose sum is 10, for which $\frac{x^2}2 + y^3$ is the smallest.
​

What you wrote -- "in such a way x^2/2 +y^3=0 is the least." -- is not possible. If x and y are positive numbers, then there is no way that $\frac{x^2}2 + y^3$ could be equal to 0.