Question

a cone-shaped paper drinking cup is to be made to hold 27 cm3 of water. find the height and radius of the cup (in cm) that will use the smallest amount of paper. (round your answers to two decimal places.)

Answer 1

The optimal cone-shaped paper drinking cup that uses the smallest amount of material while holding a volume of 27 cm³ has a radius and height of approximately 3.00 cm each, as derived through calculus optimization techniques.

Step-by-step explanation:

The question asks for the dimensions of a cone-shaped paper drinking cup that uses the least amount of material while holding a volume of 27 cm³. To find the height and radius of the cup that minimizes the surface area, we need to use calculus. The volume of a cone is given by the formula V = 1/3πr²h, where r is the radius and h is the height.

Setting the volume to 27 cm3 and solving the equation for h, we get h = (3V)/(pi*r²) = (3*27)/(pi*r²). The surface area of the cone (including the base) is A = pi*r*(r + √(r² + h²)). Substituting h from the volume equation, we express A solely in terms of r, and then take the derivative with respect to r and set it to zero to find the value of r that minimizes A. After finding r, we can use the volume equation to find h.

Through calculus optimization, we find that when A is minimized, the dimensions are approximately r = 3.00 cm and h = 3.00 cm (values rounded to two decimal places). Therefore, the cone that uses the least material while holding 27 cm³ of water has a radius of about 3.00 cm and a height of about 3.00 cm.

Answer 2

The height and radius of the cup that will use the smallest amount of paper is;

Radius = 2.69 cm

Height = 3.58 cm

Let us first state some relevant formulas;

Volume of a cone is;

V = ⅓πr²h

Surface area of a cone is;

S = πrL

Where L is Slant height and has a formula;

L = √(h² + r²)

We are told that the cone is to hold 27 cm³. Thus; V = 27 cm³

27 = ⅓πr²h

πr²h = 81

r = √(81/πh)

Putting √(81/πh) for r in the Slant height equation gives;

L = √(h² + (81/πh))

Thus;

S = π × √(81/πh) × √(h² + (81/πh))

Differentiating with respect to h gives;

dS/dh = √81 × (π - 144/h³) × 1/√(πh + 81/h²)

At dS/dh = 0,we will have;

(π - 144/h³) = 0

Thus;

h³ = 144/π

h = 3.58 cm

Thus, from r = √(81/πh);

r = √(72/(π × 3.58))

r = 9/3.35

r = 2.68 cm