Optimization box problem (How to input the information that I just solved for part C, D, & E)?

Question

zenmo · Answer

attachment

zenmo · Answer

v=Bh
v=(3-2x)(3-2x)x
$$v(x)=(9-12x+4x^2)x$$
$$v(x)=4x^3-12x^2+9$$
$$v ' (x)=12x^2-24x+9$$
$$=3(4x^2-8x+3)=0$$
$$(2x-3)(2x-1)=0$$
$$x=\frac{ 1 }{ 2} or x=\frac{ 3 }{ ? }$$, x cannot be 3/2 due to making the domain 0.
$$v(1/2)=4(1/2)^3-12(1/2)^2+9/2$$
$$=1/2-3+9/2$$
=2 cubic feet

zenmo · Answer

opps, x=1/2 or x=3/2 for near the end **

campbell_st · Answer

well you solution is correct... but the logic for selecting the value that gives the max value is incorrect... 

the domain for the question is $$0 < x < 3$$

so you need the 2nd derivative test to see which stationary point is a maximum. 

$$v"(x) = 24x - 24 $$

so if v"(1/2) < 0 then you have a max if v"(11/2) > 0 you have a min

use the same test for x = 3/2

campbell_st · Answer

try v(x) = 4x^3 - 12x^2 + 9x

use ^ for powers