# Maximum Volume...

#### rsyed5

##### New member

So, I have this question, but I have no idea what constraint is and how to find a constraint for the length, height and width.... and if i say the square wastage is x, then the width is 80-x but I don't know what the length would be with respect to x.... , and how do we determine the dimensions..?

#### Prove It

##### Well-known member
MHB Math Helper
A constraint is a restriction that is placed on an amount. Can you see any natural restrictions for the length, width and height here?

As for the length, notice that to the left, exactly half of the total 120cm is wasted, leaving 60cm, and then another x cm is removed from the right. So the length is 60 - x.

#### soroban

##### Well-known member
Hello, rsyed5!

The company asks you to generate the dimensions of the
rectangular tank that will maximize its volume.

(a) List any constraints on the length (L), width (W)
and height (H) of the tank.

(b) Determine the dimensions as exact values and also
as approximate values correct to two decimal places.
This is the correct diagram.

Code:
: - - - -  120  - - - - :
- *-------*---*-------*---* -
: |///////|///|       |///| H
: * - - - * - * - - - * - * -
: |       |   |       |   | :
80 |       |   |       |   | W
: |       |   |       |   | :
: * - - - * - * - - - * - * -
: |///////|///|       |///| H
- *-------*---*-------*---* -
: - L - : H : - L - : H :
Reading across the bottom: .$$2L + 2H \:=\:120\;\;[1]$$
. . Hence: .$$0 < L < 60$$

Reading down the right side: .$$2H + W \:=\:80\;\;[2]$$
. . Hence: .$$0 < W < 80\,\text{ and }\,0 < H < 40$$

From [1]: .$$L \:=\:60-H$$

From [2]: .$$W \:=\:80-2H$$

We have: .$$V \:=\:LWH$$

Hence: .$$V \:=\: (60-H)(80-2H)H$$

And we must maximize: .$$V \:=\:2H^3 - 200H^2 + 4800H$$