The largest possible volume of the given box is; 96.28 ft³
<h3>How to maximize volume of a box?</h3>
Let b be the length and the width of the base (length and width are the same since the base is square).
Let h be the height of the box.
The surface area of the box is;
S = b² + 4bh
We are given S = 100 ft². Thus;
b² + 4bh = 100
h = (100 - b²)/4b
Volume of the box in terms of b will be;
V(b) = b²h = b² * (100 - b²)/4b
V(b) = 25b - b³/4
The volume is maximum when dV/db = 0. Thus;
dV/db = 25 - 3b²/4
25 - 3b²/4 = 0
√(100/3) = b
b = 5.77 ft
Thus;
h = (100 - (√(100/3)²)/4(5.77)
h = 2.8885 ft
Thus;
Largest volume = [√(100/3)]² * 2.8885
Largest Volume = 96.28 ft³
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Answer:
zero
Step-by-step explanation:
Answer:
The rate of change of the volume of the cylinder at that instant = 
Step-by-step explanation:
Given:
Rate of increase of base of radius of base of cylinder = 7 mm/hr
Height of cylinder = 1.5 mm
Radius at a certain instant = 12 mm
To find rate of change of volume of cylinder at that instant.
Solution:
Let
represent radius of base of cylinder at any instant.
Rate of increase of base of radius of base of cylinder can be given as:

Volume of cylinder is given by:

Finding derivative of the Volume with respect to time.

Plugging in the values given:


Using 

(Answer)
Thus rate of change of the volume of the cylinder at that instant = 