MATH QUESTION!!

Question

From a 9​-cm by 9​-cm piece of​ cardboard, square corners are cut out so that the sides can be folded up to make a box. Complete parts a through c below.

a) Express the volume of the box as a function of the side x, in centimeters, of a cut-out square.

V=

Can't do B or C until I figure out answer a!

Answer 1

V= LWH = length times width times height

L= (9-2x) = W

H= x, 0 < x < 9/2 cm

V = (9-2x)(9-2x)x = x(9-2x)^2 = x(81-36x + 4x^2)

V = 4x^3 -36x^2 + 81x cm^3

maximum volume is when V'=0

V' = 12x^2 -72x + 81 = 0

4x^2 - 24x + 27 = 0

x^2 -6x = -27/4

complete the square, add 9 to both sides

x^2 -6x + 9 = 36/4 -27/4 = 9/4

(x-3)^2 =(3/2)^2, take the square root

x-3 = + or - 3/2

x = 3+3/2 or 3-3/2

x = 4.5 or 1.5 ignore 4.5 as that gives V=0

x= 1.5 cm gives the maximum volume

V = x(9-2x)^2 = 1.5(9-3)^2 = 1.5(6)^2 = 1.5(36) = 54 cm^3

x= 4.5 means V =0

x= 4 means V =1

x =3 means V= 27

x= 2.5 means V =40

x=2 means V = 50

x= 1.5 means V = 54

x =1 means V= 49

x= 0.5 means V =32

x = 0 means V = 0

Surface Area of the box with a closed top = S = 4(9-2x)x + 2(9-2x)^2 = 4 times the area of one side + 2 times the area of top or bottom. area of top or bottom = (9-2x)^2. area of one side

=x(9-2x)

for maximum volume the surface area =108 cm^2

S = 36x -8x^2 + 2(81-36x + 4x^2)

= 36x - 8x^2 + 162 -72x + 8x^2

S = 162 - 36x cm^2 where 0<x<9/2 cm

S(x) = S(1.5) = 162 -36(1.5) = 162 - 54 = 108 cm^2

S(x) = 162 -36x

minimum surface area is with a cube when x= 9-2x

3x = 9

x = 3 cm

max volume is 1.5 cm by 6 cm by 6 cm = 54 cm^3

which requires 108 cm^2 surface area

min surface area is 3 cm by 3 cm by 3 cm which has

volume = 27 cm^3

My guess is b and c probably involve calculating maximum volume and maybe minimum surface area?