Question 1173002
<pre>I'll answer in reverse order:</pre>
2. Express the volume of the square box in terms of x.<pre>
{{{V=l*w*h}}}
{{{V=(2x-5)*(2x-5)*(x+3)}}}
{{{200 <= V <= 810}}}
{{{200 <= l*w*h <= 810}}}
{{{200 <= b*b*h <= 810}}}
{{{200 <= (2x-5)*(2x-5)*(x+3) <= 810}}}

Multiplying that out,

{{{200 <= 4x^3 - 8x^2 -35 x + 75 <= 810}}}

We find the values of x at the endpoints 200 and 810:

{{{4x^3 - 8x^2 -35 x + 75 = 200}}} and {{{4x^3 - 8x^2 -35 x + 75 = 810}}}

{{{4x^3 - 8x^2 -35 x - 125 = 0}}} and {{{4x^3 - 8x^2 -35 x + 75 = 810}}}

Using synthetic division to factor them both, we get:

{{{(x - 5)(4x^2 + 12x + 25)=0}}} and {{{(x - 7)(4x^2 + 20x + 105)=0}}}

Those have 5 and 7 respectively as their only real solutions

The volume (2x-5)*(2x-5)*(x+3) is steadily increasing as the volume increases
from 200 cm<sup>3</sup> to 810 cm<sup>3</sup>.

So the interval for x corresponding to {{{200<=V<=810}}} is {{{5<=x<=7}}}.

</pre>1. Give the area of the base in terms of x.<pre>

Area of the square base = base<sup>2</sup> = (2x-5)<sup>2</sup>.

We build that up from this inequality:

{{{5<=x<=7}}}

We multiply all three sides by 2:

{{{2*4<=2*x<=2*7}}}

{{{8<=2x<=14}}}

We subtract 5 from all three sides:

{{{8-5<=2x-5<=14-5}}}

{{{3<=2x-5<=9}}}

We square all three sides (which keeps the same inequality since they are > 1).

{{{3^2<=(2x-5)^2<=9^2}}}

{{{9<=matrix(1,4,Area,of,square,base)<=81}}}

Edwin</pre>