Acylindrical can is to hold 20 cubic meters. the material for the top and bottom costs $10 per square meter, and material for the side costs $8 per square meter. fond the radius r and the height h of the most economical can i. e. such that the cost is minimal

The cost of the can is minimal when r=2 m and h=5 m

Step-by-step explanation:

This is an optimization problem which will be solved by the use of derivatives

We must find an expression for the total cost of the cylindrical can and then find its dimensions to make the cost minimal

Let's picture a cylinder of radius r and height h. Its volume is computed as

To make the lids we need two circle-shaped pieces of a material which costs $10 per square meter

The area of each lid is

The cost of both lids will be

The lateral side of the cylinder can be constructed with a rectangle of material which costs $8 per square meter

The rectangle has a height of h and a width equal to the length of the circumference

The cost of the side of the cylinder will be

The total cost of the can is

We know the volume of the can is cubic meters

Isolating h we have

Using this value into the total cost

Differentiating with respect to r

To find the critical point, we must set C'=0

Operating

Solving for r

r=2

And since  

h=5

Differentiating again we find

Which is always positive for r positive. It makes the critical point found a minimum

The cost of the can is minimal when r=2 m and h=5 m

right

step-by-step explanation:

answer: first option i.e. 'a' is correct.

step-by-step explanation:

since we have given that

number of minutes to wrap a small gift box = 4 minutes

we need to find the number of gifts to be wrapped in 5 minutes

since there is a direct variation, as the number of minutes will increase with   number of gifts to be wrapped.

so, let the number of gifts to be wrapped in 5 minutes be x.

so, it becomes,

hence, first option 'a'   is correct.