SOLUTION: The bottom of a box is to be a rectangle with a perimeter of 42cm. The box must have a height of 10c. what is dimensions of the box give you the maximum volume? What is the maximum

Question 825568: The bottom of a box is to be a rectangle with a perimeter of 42cm. The box must have a height of 10c. what is dimensions of the box give you the maximum volume? What is the maximum volume?
*Please show all your work so that I can see how you got your answer, Thanks!!!
Found 2 solutions by josgarithmetic, htmentor:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
bottom, w by y.
;

volume is ;

From perimeter equation,
w+y=21
w=21-y;
Substitute...
v=(21-y)y*10

Not finished, short on time here; but quadratic equation in terms of y, so you can find the maximum...
coefficient on y^2 is a negative constant, so v has a maximum.

You can take average of the roots (for when v=0) and that is the y length so you can then use it to find w. If this a Calculus problem? You could find derivative and equate to zero to find y...

Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
The volume of the box, V = l*w*h
Since the height, h is fixed at 10 cm, the problem involves varying the length and width to maximize the area.
Since the perimeter = 2(l + w) = 42 -> w = 21 - l
Now we have an expression for A in terms of the single variable, l:
A = l(21 - l) = 21l - l^2
The area will be maximized where dA/dl = 0
dA/dl = 21 - 2l = 0 -> l = 21/2 = 10.5
Therefore w = 21 - 10.5 = 10.5, i.e., the volume is maximized when the bottom is a square. The total volume, V = 10.5^2*10 = 1102.5 cm^3
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