I just don't know how to interpret this problem

[nombreuso]

Consider a paper cup in the shape of a cone obtained by rotating the line segment y = ax, 0 ≤ x ≤ r, around the y axis. For which slope a ≥ 0 will the paper cup hold the most water, assuming its surface area A is held fixed. (Use the formulas: volume V = 1 3 (area of the base)(height); A = πr^2 *√( 1 + a ^2)
Which one is the variable, x or a?
This is supossed to be a maximum and minimum problem.

 

[Dr.Peterson]

Consider a paper cup in the shape of a cone obtained by rotating the line segment y = ax, 0 ≤ x ≤ r, around the y axis. For which slope a ≥ 0 will the paper cup hold the most water, assuming its surface area A is held fixed. (Use the formulas: volume V = 1 3 (area of the base)(height); A = πr^2 *√( 1 + a ^2)
Which one is the variable, x or a?
This is supossed to be a maximum and minimum problem.

"a" is a parameter that you are to determine in order to maximize the volume. "x" is a variable in the equation of the line, which will not play an important role in your work. The important thing will be that the height is h = ar.

Do what they say: find an expression for the volume in terms of parameters a and r, and maximize that, using a as a variable and holding the surface area of the cup = A.

 

[nombreuso]

"a" is a parameter that you are to determine in order to maximize the volume. "x" is a variable in the equation of the line, which will not play an important role in your work. The important thing will be that the height is h = ar.

Do what they say: find an expression for the volume in terms of parameters a and r, and maximize that, using a as a variable and holding the surface area of the cup = A.

The thing is, since A and r are fixed, won't there be only one positive value a wich satisfies that?

 

[lev888]

A depends on a, how is it "held fixed"?

 

[Dr.Peterson]

The thing is, since A and r are fixed, won't there be only one positive value a which satisfies that?

When you don't know how a problem will work, you just have to try it and find out.

So write the equations, one for the constraint (fixed area) and one for the quantity to be maximized (volume), and then we can discuss it.

 

[nombreuso]

When you don't know how a problem will work, you just have to try it and find out.

So write the equations, one for the constraint (fixed area) and one for the quantity to be maximized (volume), and then we can discuss it.

A = πr^2 *√( 1 + a ^2) and a > 0, so a = sqrt((A^2)/(πr)^2)-1), V = π(ar)r^2 = πar^3 = πsqrt((A^2)/(πr)^2)-1)r^3

 

[Dr.Peterson]

A = πr^2 *√( 1 + a ^2) and a > 0, so a = sqrt((A^2)/(πr)^2)-1), V = π(ar)r^2 = πar^3 = πsqrt((A^2)/(πr)^2)-1)r^3

Check the power of r when you solved for a, and don't forget the factor of 1/3 in the volume of a cone.

I would eliminate r rather than a, since it is a that we want to find; we need to think of r as fixed and a variable. The work will get a little ugly at first, but will simplify nicely by the end.

 

[nombreuso]

Check the power of r when you solved for a, and don't forget the factor of 1/3 in the volume of a cone.

I would eliminate r rather than a, since it is a that we want to find; we need to think of r as fixed and a variable. The work will get a little ugly at first, but will simplify nicely by the end.

a = sqrt((A^2)/ ((π^2)(r^4)-1), V = π(ar)r^2 = (πar^3)3 = (πsqrt((A^2)/((π^2)(r^4)-1)r^3) What do you mean by 'eliminate'.There's still only one possible value for a.

 

[Dr.Peterson]

a = sqrt((A^2)/ ((π^2)(r^4)-1), V = π(ar)r^2 = (πar^3)3 = (πsqrt((A^2)/((π^2)(r^4)-1)r^3) What do you mean by 'eliminate'.There's still only one possible value for a.

I see your issue now. You have to consider r variable, as I did in my work without connecting it to what I was saying. The constraint relates the variables a and r, so you solve for r in terms of a in order to express V as a function only of a.