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Using differential calculus, maximize the volume of a box made of cardboard (top is open) as shown in Figure A. 15, subject to the following constraints. 1. Allowable area of the cardboard is equal to 40 in^2. 2. The length of the box L is equal to its width W.

Sagot :

The maximum volume of the box is 40√(10/27) cu in.

Here we see that volume is to be maximized

The surface area of the box is 40 sq in

Since the top lid is open, the surface area will be

lb + 2lh + 2bh = 40

Now, the length is equal to the breadth.

Let them be x in

Hence,

x² + 2xh + 2xh = 40

or, 4xh = 40 - x²

or, h = 10/x - x/4

Let f(x) = volume of the box

= lbh

Hence,

f(x) = x²(10/x - x/4)

= 10x - x³/4

differentiating with respect to x and equating it to 0 gives us

f'(x) = 10 - 3x²/4 = 0

or, 3x²/4 = 10

or, x² = 40/3

Hence x will be equal to 2√(10/3)

Now to check whether this value of x will give us the max volume, we will find

f"(2√(10/3))

f"(x) = -3x/2

hence,

f"(2√(10/3)) = -3√(10/3)

Since the above value is negative, volume is maximum for x = 2√(10/3)

Hence volume

= 10 X 2√(10/3)  -  [2√(10/3)]³/4

= 2√(10/3) [10 - 10/3]

= 2√(10/3) X 20/3

= 40√(10/27) cu in

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