Question

A box with a square base and open top must have a volume of 4,000 cm. Find the dimensions of the box that minimize the amount of material used. 

Answer 1

volume V2 = 4,000 2h = 4,000 h = 4000 SIA(S): 2742 h = 27°+44(4000) = 28+ 16000 Act? = 42– 16.000 C-p.when A=0 = 4*-16000 47=16000 232 4000 Y-1014 *3 A'<»= 4- 16000(-223) = 4+39.000 ACTO 322 47 32000 30 4000 A man at az 10 VG side bare (2) = 1084 - height Ch?= 4,000 = 4,000 1000 (443) 2 sene ya Foto 2 21-43214

Answer 2

SOLUTION :

Let the square base sides be x cm and height be y cm. Top of the box is open.

So,

Volume, V = x*x*y = x^2 y cm^3

=> 4000 = x^2 y 

=> y = 4000/x^2 

Area of base and sides, A = x^2 + 4x y 

=> A = x^2 + 4x * 4000/x^2 = x^2 + 16000/x

For minimum amount of material, area A of base and sides should be minimum.

For minimum area A, 

dA/dx = 0

=> d/dx (x^2 + 16000/x) = 0

=> 2x - 16000/x^2 = 0

=> 2x = 160000/x^2

=> 2x^3 = 16000

=> x = cube root (16000/2) = cube root (8000) = 20 cm

A(20) = (20)^2 + 16000/20 = 1200 cm^2 .

A(19) = 19^2 + 16000/19 = 1203.11 cm^2

A(21) = 21^2 + 16000/21 = 1202.90 cm^2

Hence, A is minimum at x = 20 cm.

=> y(20)= 4000/20^2 = 10 cm.

Hence ,for minimum material to form the top open box of square base and suitable height to have volume of 4000 cm^3,

Sides of square base = 20 cm.  (ANSWER)

Height = 10 cm (ANSWER).