You are asked to build an open cylindrical can (i. e. no top) that will hold 364.5 cubic inches. To do this, you will cut its bottom from a square of metal and form its curved side by bending a rectangular sheet of metal.(a) Express the total amount of material required for the square and the rectangle in terms of r. A(r) = (b) Find the radius and height of the can that will minimize the total amount of material required. r = h =

  (a) A(r) = 4r² +729/r

  (b) r = 4.5 inches; h = 5.73 inches

Step-by-step explanation:

(a) For a given radius, the volume is given by the formula ...

  V = πr²h

so the height is ...

  h = V/(πr²)

Then the area of the rectangle required to form the curved side is ...

  A = 2πrh = (2πr)(V/(πr²)) = 2V/r

The area required for the bottom is a square that is 2r on each side, so is ...

  A = (2r)² = 4r²

The total area of required material is ...

  A(r) = 4r² +2V/r

  A(r) = 4r² +729/r

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(b) The material will be minimized when the derivative of A(r) with respect to r is zero:

  A' = 0 = 8r -2V/r²

  2V/r² = 8r

  V/4 = r³

  r = ∛(V/4) = ∛(364.5/4) ≈ 4.5 . . . inches

  h = (364.5)/(π·4.5²) = 18/π ≈ 5.73 . . . inches

Material will be minimized for a radius of 4.5 inches, and a height of 5.73 inches.

d) 30 units^2

area of rectangle = l × b

we know,

length = 10 unit

breadth = 3 unit

applying the formula,

= l × b

= 10 × 3

= 30 unit^2

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