# max/min

• Mar 9th 2013, 01:06 PM
apatite
max/min
A rectangular tank with a bottom and sided but but no top is to have volume 500 cubic feet. Determine the

dimensions (length, width, height) with the smallest possible surface area.
• Mar 9th 2013, 03:13 PM
ILikeSerena
Re: max/min
Quote:

Originally Posted by apatite
A rectangular tank with a bottom and sided but but no top is to have volume 500 cubic feet. Determine the

dimensions (length, width, height) with the smallest possible surface area.

Hi apatite! :)

For a minimal surface area, the top of the tank will be square.
Let's say the side of this square is x.
And let h be the height of the tank.

Then the volume is $x \times x \times h = 500$ and the surface area is $A = x \times x + 4 x \times h$.

Can you combine these equations to find A as a function of x?
Afterwards its derivative needs to be zero, which will give you x.
• Mar 12th 2013, 06:54 AM
apatite
Re: max/min
Quote:

Originally Posted by ILikeSerena
Hi apatite! :)

For a minimal surface area, the top of the tank will be square.
Let's say the side of this square is x.
And let h be the height of the tank.

Then the volume is $x \times x \times h = 500$ and the surface area is $A = x \times x + 4 x \times h$.

Can you combine these equations to find A as a function of x?
Afterwards its derivative needs to be zero, which will give you x.

derivative becomes zero. I think your answer is wrong bud
• Mar 12th 2013, 07:05 AM
ILikeSerena
Re: max/min
Quote:

Originally Posted by apatite
derivative becomes zero. I think your answer is wrong bud

Neh, it only means you did not combine the equations as I suggested... bud!
• Mar 12th 2013, 07:17 AM
apatite
Re: max/min
Quote:

Originally Posted by ILikeSerena
Neh, it only means you did not combine the equations as I suggested... bud!

what do you mean? h= 500/x^2 and then you substitute that in the second equation and set derivative equal zero?
• Mar 12th 2013, 07:19 AM
ILikeSerena
Re: max/min
Quote:

Originally Posted by apatite
what do you mean? h= 500/x^2 and then you substitute that in the second equation and set derivative equal zero?

Yep!
• Mar 12th 2013, 07:24 AM
apatite
Re: max/min
Quote:

Originally Posted by ILikeSerena
Yep!

A= 5x^2 * 500/x^2 the x^2 cancels out. That's what I don't get.
• Mar 12th 2013, 07:28 AM
ILikeSerena
Re: max/min
Quote:

Originally Posted by ILikeSerena
Hi apatite! :)

For a minimal surface area, the top of the tank will be square.
Let's say the side of this square is x.
And let h be the height of the tank.

Then the volume is $x \times x \times h = 500$ and the surface area is $A = x \times x + 4 x \times h$.

Can you combine these equations to find A as a function of x?
Afterwards its derivative needs to be zero, which will give you x.

Quote:

Originally Posted by apatite
A= 5x^2 * 500/x^2 the x^2 cancels out. That's what I don't get.

Looks like you did not substitute correctly.

$x \times x \times h = 500$

$h = \frac {500}{x \times x} \qquad (1)$

Substitute (1) into the formula for the area:

$A = x \times x + 4 x \times h$

$A = x \times x + 4 x \times \left(\frac {500}{x \times x}\right)$
• Mar 12th 2013, 07:46 AM
apatite
Re: max/min
Quote:

Originally Posted by ILikeSerena
Looks like you did not substitute correctly.

$x \times x \times h = 500$

$h = \frac {500}{x \times x} \qquad (1)$

Substitute (1) into the formula for the area:

$A = x \times x + 4 x \times h$

$A = x \times x + 4 x \times \left(\frac {500}{x \times x}\right)$

ooo so A= (x^2+4x) * (5oo/x^2). Thank you for your help ;)
• Mar 12th 2013, 08:00 AM
ILikeSerena
Re: max/min
Quote:

Originally Posted by ILikeSerena
Looks like you did not substitute correctly.
$x \times x \times h = 500$

$h = \frac {500}{x \times x} \qquad (1)$

Substitute (1) into the formula for the area:
$A = x \times x + 4 x \times h$

$A = x \times x + 4 x \times \left(\frac {500}{x \times x}\right)$

Quote:

Originally Posted by apatite
ooo so A= (x^2+4x) * (5oo/x^2). Thank you for your help ;)

Not quite.
Multiplication has a higher priority than addition.
$A = x \times x + 4 x \times \left(\frac {500}{x \times x}\right)$

$A = x \times x + \left(\frac {4 x \times 500}{x \times x}\right)$

$A = x^2 + \frac {2000}{x}$