# Math Help - max/min

1. ## 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.

2. ## Re: max/min

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.

3. ## Re: max/min

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.

4. ## Re: max/min

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

5. ## Re: max/min

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?

6. ## Re: max/min

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!

7. ## Re: max/min

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

8. ## Re: max/min

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.
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:

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

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

9. ## Re: max/min

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

10. ## Re: max/min

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

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

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

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