Closed Calculus

[Ferocity]

Pa help dito mga sir medyo nahirapan ako e haha.
"The volume of an open box with a square base is 4000cm^3. Find the dimensions of the box if the material used to make the box is minimum?"

 

[conviction]

Let the side of the square base = s.
Let the height of the box = h.
Then the volume of the box, V = s²h.

The area of the base = s².
There are 4 sides, each side has an area = s.h.
So total area of 4 sides = 4sh.
So, total area, 4 sides + base = s² + 4sh.
Let A = s² + 4sh.

So, we have: V = 4000 = s²h, and A = s² + 4sh.
Therefore, h = 4000/s²; so A = s² + 4s.(4000/s²) = s² + 16,000/s,
i.e.: A = s² + 16,000/s.

We want to find s for A = a minimum.
That will be when dA/ds = 0.
Since: A = s² + 16,000/s
dA/ds = 2s - 16,000/s².
For dA/ds = 0, 2s = 16,000/s²,
so: 2s^3 = 16,000; s = 20.

We know that: V = 4000 = s²h.
So, 4000 = 20²h = 400h, so h = 10.

Hence, the square base is (20 X 20) cm.; each side has height 10 cm. & width 20 cm.
So, total material used = (20 X 20) + 4(20 X 10) = 1200 sq. cm.

source: yahoo
HTH,
Skywave.

 

[mito02]

Source: google

 

[Deleted member 622421]

Given Volume = 4000 cm^3 = (x^2)

So y = 4000/(x^2)

Substitute the y into the area function.

Area = (x^2) + 4x [ 4000 / (x^2) ]
Area = (x^2) + 16 000/x

I use differentiation method to find the minimum dimension of the box.

Let y = (x^2) + 16 000/x
dy/dx = 2x + (- 16 000 x^ -2)
dy/dx = 2x - 16 000/(x^2)

Differentiate dy/dx again, d^2y/dx^2 = 2 + 32 000 x^ -3
So Second differentiation = 2 + 32 000/(x^3)

When dy/dx = 0, 2x - 16 000/(x^2) = 0
2x^3 -16000 = 0
2X^3 = 16000
x^3 = 8000
x = 20

When x = 20, second differentiation = 2 + 32 000/(2000^3)
second differentiation = 6 (positive means this is minimum)

Sometimes we get 2 answers for x, that is why we use second differentiation to determine which x is maximum or minimum.

Therefore, when x = 20, y = 4000/20^2
y = 10
So length/width = 20 cm, height = 10 cm.

 

[Ferocity]

Pa help dito mga sir medyo nahirapan ako e haha.
"The volume of an open box with a square base is 4000cm^3. Find the dimensions of the box if the material used to make the box is minimum?"

Salamat sa mga sagot.