Answer:
Possible expressions for the radius of the can and the depth of the paint in the can are [tex]r = \sqrt{9\cdot x^{2}+24\cdot x+16}[/tex] and [tex]h = 2\cdot x[/tex], respectively.
Step-by-step explanation:
Let be the initial volumes of the initial cans represented by these expressions:
[tex]V_{1} = (8\cdot x^{3}+31\cdot x^{2}+32\cdot x)\cdot \pi[/tex] (1)
[tex]V_{2} = (10\cdot x^{3}+17\cdot x^{2})\cdot \pi[/tex] (2)
The resulting volume of the paint can is the sum of the two functions:
[tex]V_{3} = (18\cdot x^{3}+48\cdot x^{2}+32\cdot x)\cdot \pi[/tex] (3)
Then, we proceed to factor the polynomial:
[tex]V_{3} = 2\cdot (9\cdot x^{2}+24\cdot x +16)\cdot x \cdot \pi[/tex]
[tex]V_{3} = \pi\cdot (9\cdot x^{2}+24\cdot x + 16)\cdot (2\cdot x)[/tex] (3b)
By direct comparison with the volume formula for the cylinder we have the following expressions:
[tex]r^{2} = 9\cdot x^{2}+24\cdot x + 16[/tex]
[tex]r = \sqrt{9\cdot x^{2}+24\cdot x+16}[/tex]
[tex]h = 2\cdot x[/tex]
Possible expressions for the radius of the can and the depth of the paint in the can are [tex]r = \sqrt{9\cdot x^{2}+24\cdot x+16}[/tex] and [tex]h = 2\cdot x[/tex], respectively.
