SOLUTION: The volume of a square box is 200 cm3 ≤ 𝑉 ≤ 810 cm3 . If the height is (x + 3) cm and the side of the square base is (2x - 5) cm. 1. Give the area of the base in term

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: The volume of a square box is 200 cm3 ≤ 𝑉 ≤ 810 cm3 . If the height is (x + 3) cm and the side of the square base is (2x - 5) cm. 1. Give the area of the base in term      Log On

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Question 1173002: The volume of a square box is 200 cm3 ≤ 𝑉 ≤ 810 cm3
. If the height is
(x + 3) cm and the side of the square base is (2x - 5) cm.
1. Give the area of the base in terms of x.
2. Express the volume of the square box in terms of x.

Found 2 solutions by Edwin McCravy, mccravyedwin:
Answer by Edwin McCravy(20054) About Me  (Show Source):
Answer by mccravyedwin(407) About Me  (Show Source):
You can put this solution on YOUR website!
I'll answer in reverse order:

2. Express the volume of the square box in terms of x.
V=l%2Aw%2Ah
V=%282x-5%29%2A%282x-5%29%2A%28x%2B3%29
200+%3C=+V+%3C=+810
200+%3C=+l%2Aw%2Ah+%3C=+810
200+%3C=+b%2Ab%2Ah+%3C=+810
200+%3C=+%282x-5%29%2A%282x-5%29%2A%28x%2B3%29+%3C=+810

Multiplying that out,

200+%3C=+4x%5E3+-+8x%5E2+-35+x+%2B+75+%3C=+810

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

4x%5E3+-+8x%5E2+-35+x+%2B+75+=+200 and 4x%5E3+-+8x%5E2+-35+x+%2B+75+=+810

4x%5E3+-+8x%5E2+-35+x+-+125+=+0 and 4x%5E3+-+8x%5E2+-35+x+%2B+75+=+810

Using synthetic division to factor them both, we get:

%28x+-+5%29%284x%5E2+%2B+12x+%2B+25%29=0 and %28x+-+7%29%284x%5E2+%2B+20x+%2B+105%29=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 cm3 to 810 cm3.

So the interval for x corresponding to 200%3C=V%3C=810 is 5%3C=x%3C=7.

1. Give the area of the base in terms of x.

Area of the square base = base2 = (2x-5)2.

We build that up from this inequality:

5%3C=x%3C=7

We multiply all three sides by 2:

2%2A4%3C=2%2Ax%3C=2%2A7

8%3C=2x%3C=14

We subtract 5 from all three sides:

8-5%3C=2x-5%3C=14-5

3%3C=2x-5%3C=9

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

3%5E2%3C=%282x-5%29%5E2%3C=9%5E2

9%3C=matrix%281%2C4%2CArea%2Cof%2Csquare%2Cbase%29%3C=81

Edwin