MHB Dimensions of a rectangular prism

AI Thread Summary
The volume of a rectangular prism is represented by the polynomial V(x) = 2x^3 + 9x^2 + 4x - 15, with the depth given as (x-1) feet and the length as 13 feet. By using synthetic division, it is established that V(x) can be factored into (x-1)(2x+5)(x+3). To find the value of x that makes the largest factor equal to 13, it is determined that x=4 satisfies this condition. Consequently, the dimensions of the tank are 3 ft, 7 ft, and 13 ft.
Madds
Messages
2
Reaction score
0
The volume of a rectangular prism can be represented by the polynomial
V(x)=2x^2+9x^2+4x-15
a. The depth of the tank is (x-1) feet. The length is 13 feet. Assume the length is the greatest dimension. Which linear factor represents the 13 ft?This is probably a really easy question but I am so confused reading it, I really need help on how to do these kinds of problems.
 
Mathematics news on Phys.org
Re: Help with an equation problem

I am assuming the volume is the cubic polynomial:

$$V(x)=2x^3+9x^2+4x-15$$

We are told the depth is $x-1$, so we know $1$ is a zero of the polynomial...so let's use synthetic division:

$$\begin{array}{c|rr}& 2 & 9 & 4 & -15 \\ 1 & & 2 & 11 & 15 \\ \hline & 2 & 11 & 15 & 0 \end{array}$$

So, we now know:

$$V(x)=(x-1)\left(2x^2+11x+15\right)$$

Now we need to factor the quadratic factor...

$$V(x)=(x-1)(2x+5)(x+3)$$

What value of $x$ makes the largest factor equal to 13?
 
Re: Help with an equation problem

As a followup, we can determine which value of $x$ makes the largest factor 13 by looking at the following graph:

View attachment 7450

We can easily see that when $x=4$, the largest linear factor is $2x+5=13$. The dimensions of the tank are:

$$3\text{ ft}\times7\text{ ft}\times13\text{ ft}$$
 

Attachments

  • prismfactors.png
    prismfactors.png
    17.8 KB · Views: 100
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top