Minimizing surfave area for a fixed volume, finding max area (Calculus)

[EETman]

I will list off the problems exactly how they read then provide my work up to this point:
1.
A box has a square base, x*x, height h, and an open top.
What is the ratio h/x that would minimize the total surface area for a fixed volume V?

so far i have determined:
b=x*x
h=h
v=b*h

also, surface are is SA=4hx+2b.

In previous problems in class, we would normally take the derivative at this point, but this problem seems
very rudementary for that to be necessary, what am i missing here?

Second Problem:
a rectangl has one side on the x-axis, one side on the y-axis, one
vertex on the origin and one vertex on the curve y=e^-x , x>0. find the maximum area possible.

The curve simply goes from infinity down to a horizontal asymptote on the x-axis,
seems as if the volume is infinitely large? some advice on where to start would be greatly helpful.

These 2 problems are the "odd ones out" on my homework that has otherwise been flawlessly completed
i would appreciate any and all advice as to where i have went wrong.

 

[Romsek]

I will list off the problems exactly how they read then provide my work up to this point:
1.
A box has a square base, x*x, height h, and an open top.
What is the ratio h/x that would minimize the total surface area for a fixed volume V?

so far i have determined:
b=x*x
h=h
v=b*h

also, surface are is SA=4hx+2b.

In previous problems in class, we would normally take the derivative at this point, but this problem seems
very rudementary for that to be necessary, what am i missing here?

SA = 4hx + b as it's an open top.

I'm going to assume you aren't learning Lagrange multipliers.

Completing this is pretty straightforward.

You are maximizing a function subject to a constraint. Use the constraint to reduce your function of 2 vars, to a function of 1 var. Differentiate and set equal to 0. Solve.

In this case you have constant volume V. so V = x² h , h = V/x²

Plug that into your surface area to obtain SA = 4 (V/x²) x + x²

Differentiate and set to 0 and solve. I leave that to you.

Then when you have x. h/x = (V/x²)/x = V/x³

 

[Romsek]

I will list off the problems exactly how they read then provide my work up to this point:

Second Problem:
a rectangl has one side on the x-axis, one side on the y-axis, one
vertex on the origin and one vertex on the curve y=e^-x , x>0. find the maximum area possible.

The curve simply goes from infinity down to a horizontal asymptote on the x-axis,
seems as if the volume is infinitely large? some advice on where to start would be greatly helpful.

These 2 problems are the "odd ones out" on my homework that has otherwise been flawlessly completed
i would appreciate any and all advice as to where i have went wrong.

You should just work this one, it's not hard.

You have a rectangle one side being x long, the other side being e^-x long, so the area A=x e^-x.

You know how to maximize that area.

 

[HallsofIvy]

I will list off the problems exactly how they read then provide my work up to this point:
1.
A box has a square base, x*x, height h, and an open top.
What is the ratio h/x that would minimize the total surface area for a fixed volume V?

so far i have determined:
b=x*x
h=h
v=b*h

also, surface are is SA=4hx+2b.

The surface area is \(\displaystyle SA= 4hx+ x^2\). I see now reason to add "b" here. We are also give that \(\displaystyle x^2h= V\) so that \(\displaystyle h= \frac{V}{x^2}\). Replacing h with that in SA gives \(\displaystyle SA= \frac{4x}{x^2}+ x^2= \frac{4}{x}+ x^2\).

In previous problems in class, we would normally take the derivative at this point, but this problem seems
very rudementary for that to be necessary, what am i missing here?

Second Problem:
a rectangl has one side on the x-axis, one side on the y-axis, one
vertex on the origin and one vertex on the curve y=e^-x , x>0. find the maximum area possible.

The curve simply goes from infinity down to a horizontal asymptote on the x-axis,
seems as if the volume is infinitely large? some advice on where to start would be greatly helpful.

If the base has length "x" then the height is \(\displaystyle y= e^{-x}\) so the area is \(\displaystyle A= xe^{-x}\)

These 2 problems are the "odd ones out" on my homework that has otherwise been flawlessly completed
i would appreciate any and all advice as to where i have went wrong.