Given: A metal box with a square base and vertical sides is to contain 1024 cm³, the material for the top and bottom costs Rs 5/cm² and the material for the sides costs Rs 2.50/cm².

To find: the least cost of the box

Let the side of the square be x cm and

Let the vertical side of the metal box be y cm.

Then as per the given criteria the volume of the metal box with square base will be,

V=base × height

Here base is square, so volume becomes

V=x²y

This is equal to 1024 cm³. Hence volume becomes

1024cm³=x²y

Now, we will find the total area of the metal box.

As it is given the base is square, that means top is also a square, so

Area of top and bottom = 2x²cm²

Given the material for the top and bottom costs Rs 5/cm², so the material cost for top and bottom becomes

Cost of top and bottom =Rs. 5(2x²)

Now the vertical side of metal box is y cm, so

Area of one side of the metal box = xy cm

Now the metal box has 4 sides, so

Area of all the sides of the metal box = 4xy

Given the material for the sides costs Rs 2.50/cm²

∴ Cost of all the sides of the metal box =Rs. 2.50(4xy)

So the total area of the metal box is

A=2x²+4xy

And the total cost of the metal box is

C=5(2x²)+ 2.50(4xy)

Substituting the value of y from equation (i) in the above equation, we get

Now differentiating both sides with respect to x, we get

Applying differentiation rule of sum, we get

Now applying the derivative, we get

Now we will apply second derivative test to find out the minimum value of x, so for that let C=0, so equating above equation with 0, we get

⇒x³=512

Solving this we get

⇒ x=8

Differentiating equation (ii) again with respect to x, we get

Applying differentiation rule of sum, we get

At x=8, the above equation becomes,

Now at x=8, C’(8)=0 and C’’(8)>0, so as per the second derivative test, x is a point of local minima and C(8) will be minimum value of C.

Hence least cost becomes

⇒C_x=8=640+1280=Rs.1920

Hence the least cost of the metal box is Rs. 1920