Manipulating Expressions

xobreanne

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Nov 30, 2006
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Amanda is building a wooden rectangular toy storage box for Student Council to be raffled after the Santa Claus Parade. The box will have an open top and a volume of 9m^3. For design purposes, Amanda would like the length of the bas to be triple its width. Thick wood for the base costs $8/m^2 and thinner wood for the sides cost $5/m^2

I have already come up with the equation for the cost of wood as a function of the width of the base, and that is: 24w^2 + 120/w

the questions are:

a) If amanda can afford to spend $144 to make the toy storage box, what must the dimensions of the box be to keep the volume 9m^3.

b) What dimensions do you recommend that Amanda use? Justify your answer.
 
a. set 144 equal to your cost function ... solve for w, determine L and h.

24w^2 + 120/w = 144
24w^3 + 120 = 144w
w^3 + 5 = 6w
w = 1

b. find the minimum cost for the box by taking the derivative of the cost function, set it equal to 0, and determining the width for the cheapest box, then the other two dimensions. the absolute minimum cost is cheaper than $144.
 
how would i find the derivative of that, I cant think of how to do it.

you are in a calculus class, correct?

C = 24w<sup>2</sup> + 120/w

C = 24w<sup>2</sup> + 120w<sup>-1</sup>

now ... use the power rule for derivatives to find dC/dw.
 
i ended up getting:

0=48w - 120/w^2
w^2 = 48w - 120
w^2 - 48w = -120
w^2 + w = 2.5
w^3 = 2.5
w = 1.36

does that look right?
 
you started out fine ... and miraculously got the correct solution, but your algebra steps are incorrect.

0=48w - 120/w^2
w^2 = 48w - 120 this step is not correct
w^2 - 48w = -120
w^2 + w = 2.5
w^3 = 2.5 since when does w<sup>2</sup> + w = w<sup>3</sup>?
w = 1.36

48w - 120/w<sup>2</sup> = 0

2w - 5/w<sup>2</sup> = 0

2w = 5/w<sup>2</sup>

w<sup>3</sup> = 5/2

w = (5/2)<sup>1/3</sup> = approx 1.36
 
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