Would you please help me to answer this problem??? I also need the solution?

Question

-Application of maxima and minima
1) A closed box with square base is to have a volume of 2000 in. The material for the top and the bottom of the box is to cost 3 cents per sq.in and the material for the sides is to cost 1.5 cents/in². Estimate to analytically the dimension of the box so that the total cost of the material is least.

Answer 1

Let x = length of one side of the square base in inches.
Let h = height of the box in inches.

Solve for h, in terms of x:
x * x * h = 2000
h = 2000/x²

You have a top and bottom with an area of x²
So the cost would be 2x² * 3 cents for those.
Cost of top and bottom --> 6x²

And you have 4 sides with areas of hx.
= 4 * x * 2000/x² = 8000/x
So the cost would be 1.5 * 8000/x cents for the sides.
Cost of sides --> 12000/x

Your function for the cost is therefore:
f(x) = 6x² + 12000/x

This is the function you want to minimize, so take the derivative, solve for f'(x) = 0.

f(x) = 6x² + 12000x^-1
f'(x) = 12x - 12000x^-2
f'(x) = 12x - 12000/x²

Set this to 0:
0 = 12x - 12000/x²
12x = 12000/x²

Divide both sides by 12:
x = 1000/x²

Multiply both sides by x²:
x^3 = 1000

Take the cube root of both sides:
x = ³√1000
x = 10

Now solve for the height:
h = 2000/x²
h = 2000/100
h = 20

The dimensions of the box are:
10 inches (length) x 10 inches (width) x 20 inches (height)

Answer 2

OK, I think this is it...

Let V = volume, C = total cost.

V = LWH
Since the bottom is a square, L=W. So,
V = L²H = 2000

C = 2L² (3 cents) + 4LH(1.5 cents)
(The 2 is for top + bottom. 4 for 4 sides)
(substitute H=2000/ L²)
C = 6L² + 6L(2000/L²)
= 6L² + 12000/L

dC/dL = 12L + (-12000)/L²
Minima is where dC/dL = 0.
12L³ - 12000 = 0,
which is L = 10.

So, your dimensions are 10 x 10 x 20.

Cost = 2(10)(10)(3) + 4(10)(20)(1.5)
= 1800 cents.

Edit: oops.

Answer 3

given V=2000

square base means that the area of both top and base are

x²+x² = 2x²

the area of the 4 sides = 4xy (assuming that the hight of the box is y)

so the area of all sides of the box = 2x² + 4xy

cost = 3(2x² ) + 1.5(4xy)

= 6x² + 6xy .......(1)

since V=2000
V= x*x*y =2000
x²y = 2000
y = 2000/x²
replace ( y ) in ...(1)

cost = 6x² + 6x(2000/x² )
= 6x² + 12000/x
cَ = 12x + -12000/x²

12x + -12000/x² = 0
12x = 12000/x²
x= 10

cً > 0 so it's min value when x=10

y = 2000/x²

y =2000/100

y= 20