express cost as function of width, find best dimensions

[henry200]

i just wanted to know if this was correct, here's the question:

There's an open top box with the volume of 9m^3 and the length of its base is triple its width. The thick wood used for the base is $8/m^2 and thinner wood for sides is $5/m^2.

(a) Express the cost of wood as a function of the width of the base.

(b) If you had $144 to spend, what dimensions would the box be to keep the volume at 9m^3.

(c) What dimensions would be best?

well i have no clue for the last question (c) but for question (a) i got f(w)=24w^2 + 120/w
for (b) i got V=(area)(height)=24(w^3+5+6w).....is any of this right?

 

[stapel]

well i have no clue for the last question (c)

Hint: You probably want to spend less, rather than more.

for question (a) i got f(w)=24w^2 + 120/w
for (b) i got V=(area)(height)=24(w^3+5+6w).....is any of this right?

Note: Since you haven't shown any work, it is not possible to trouble-shoot any errors. Sorry.

a) I get the same cost function: f(w) = 24w² + 120/w :wink:

b) I'm not sure what you're doing here...? You are given that f(x) is the cost f of the box, given width w and volume 9. Where is this new equation coming from? How does it account for the given conditions? What was your reasoning? :shock:

You have volume f for width w (and thus length 3w and height 3/w²). So plug the given value in for the volume, and solve for the width. :idea:

Eliz.

 

[henry200]

when trying to solve for w, i get 24w(w^2-6)+120=0....i don't know where to go from here, what happens to the 120? without it i would get w=0 and w=2.45

 

[stapel]

when trying to solve for w, i get 24w(w^2-6)+120=0....i don't know where to go from here, what happens to the 120? without it i would get w=0 and w=2.45

You can't just discard bits of an equation you don't like. You have to solve the equation as it is; the solution to a different equation is highly unlikely to be valid for the given equation. :shock:

I'm not sure where you're getting "24w(w² - 6) + 120 = 0" from...? I'm guessing you're referring to part (b), but the above equation suggests that you're using zero for the cost, and that neither makes sense nor matches what the exercise said...?

Please reply with clarification. Thank you.

Eliz.

 

[henry200]

i substituted 144 for f(w) to get 144=24w^2+120/w. i then moved the 144 over to get 0=24w^2+120/w-144. i then multiplied each term by w to get 0=24w^3+120-144w...thats how i got 24w(w^2-6)+120=0

 

[stapel]

i substituted 144 for f(w) to get 144=24w^2+120/w. i then moved the 144 over to get 0=24w^2+120/w-144. i then multiplied each term by w to get 0=24w^3+120-144w...thats how i got 24w(w^2-6)+120=0

Why did you "factor" the two terms like that...? Does that fit some formula you were given...?

In standard formatting, you have:

. . . . .24w³ - 144w + 120 = 0

Divide through by the common factor:

. . . . .w³ - 6w + 5 = 0

From the Rational Roots Test, you know that the only possible "nice" roots are -1, +1, -5, and +5. Use synthetic division to find the one rational zero, and then use the Quadratic Formula to find the other zeroes. Since one irrational zero is negative, you can immediately discard this. See what you can do with the other two solutions.

Eliz.

 

[henry200]

if you divided through by a common factor isn't that factor supposed to be there, like...24(w^3-6w+5)=0 and not just w^3-6w+5=0?

 

[stapel]

If I'd factored, then the factor would remain, out front. Instead, I divided.

Eliz.

 

[henry200]

i got w=1 because from rational roots, it was the only one that equated to zero when placed in the formula