Question

this question has two parts. first, answer part a. then, answer part b.
part a
shipping a shipping company will ship a package for $7.50 when the volume is no more than 15,000 cubic centimeters. grace needs to ship a package that is 3x - 5 centimeters long, 2x centimeters wide, and x + 20 centimeters tall.
a. write a polynomial equation to represent the situation if grace plans to spend a maximum of $7.50.
\boxed{}x^3+\boxed{}x^2+\boxed{}x = \boxed{}
part b
b. write and solve a system of equations.
y = \boxed{}x^3+\boxed{}x^2+\boxed{}x, y = \boxed{}; x = \boxed{} cm
c. what should the dimensions of the package be to have the maximum volume?
\boxed{} cm by \boxed{} cm by \boxed{} cm

Explanation:

Step1: Calculate package volume

Volume $V = (3x-5)(2x)(x+20)$
First expand $(3x-5)(2x) = 6x^2 -10x$
Then multiply by $(x+20)$:
$$(6x^2 -10x)(x+20) = 6x^3 + 120x^2 -10x^2 -200x = 6x^3 + 110x^2 -200x$$
Set equal to max volume 15000:
$6x^3 + 110x^2 -200x = 15000$

Step2: Define system for Part B

Set $y$ equal to volume polynomial and max volume:
$y = 6x^3 + 110x^2 -200x$, $y=15000$
Simplify the equation:
$6x^3 + 110x^2 -200x -15000=0$
Divide by 2: $3x^3 + 55x^2 -100x -7500=0$
Factor by grouping:
$x^2(3x+55) - 100(3x+55)=0$
$(x^2-100)(3x+55)=0$
Solutions: $x=10, x=-10, x=-\frac{55}{3}$
Only positive $x=10$ is valid.

Step3: Find max volume dimensions

Substitute $x=10$:
Length: $3(10)-5=25$ cm
Width: $2(10)=20$ cm
Height: $10+20=30$ cm

Answer:

Part A

$6x^3 + 110x^2 -200x = 15000$
(Blanks: 6, 110, -200, 15000)

Part B

b. $y = 6x^3 + 110x^2 -200x$, $y=15000$; $x=10$
(Blanks: 6, 110, -200, 15000, 10)
c. 25 cm by 20 cm by 30 cm