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Mathematics 22 Online
OpenStudy (anonymous):

Help? A rectangular storage container with an open top is to have a volume of 18 cubic meters. The length of its base is twice the width. Material for the base costs 13 dollars per square meter. Material for the sides costs 7 dollars per square meter. Find the cost of materials for the cheapest such container.

OpenStudy (anonymous):

I alread have w^3=378/52 which is 7.27 What should I do step by step next????

OpenStudy (anonymous):

i think it's w^3=252/54 to match your answer?

ganeshie8 (ganeshie8):

hw did u get w^3 = 378/52 ?

OpenStudy (anonymous):

could be wrong though

OpenStudy (anonymous):

i'm rushing it's due in like 33 mins hehe

ganeshie8 (ganeshie8):

im aslo getting w^3 = 378/54,

ganeshie8 (ganeshie8):

when u let f'(w) = 0

OpenStudy (anonymous):

yes that's right i just recheck it

OpenStudy (anonymous):

so what do i need to do next?

ganeshie8 (ganeshie8):

find value 'w', and substitute in cost function ?

ganeshie8 (ganeshie8):

w = cuberoot(378/52)

OpenStudy (anonymous):

yeah= 8.088453308456256

ganeshie8 (ganeshie8):

whats 8.088 ?

ganeshie8 (ganeshie8):

wolfram says cost is ~ 293 http://www.wolframalpha.com/input/?i=26*%281.937%29%5E2+%2B+378%2F%281.937%29

OpenStudy (anonymous):

\[\text{The volume of a rectangular prism is as follows.}\\ \ \ \ V = lwh\\ \text{Given as follows.}\\ \ \ \ V = 18\\ \ \ \ l = 2w\\ \text{Derive for the area of the base and sides, respectively.}\\ \ \ \ A_1 = lw\\ \ \ \ A_2 = 2h(w + l)\\ \text{Let's try to reduce our volume equation.}\\ \ \ \ 18 = 2w^2h\\ \ \ \ h = 9w^{-2}\\ \text{Derive the cost function.}\\ \ \ \ C(w) = 13A1 + 7A2\\ \ \ \ \ \ \ \ \ \ \ \ = 13lw + 14h(w+l)\\ \ \ \ \ \ \ \ \ \ \ \ = 26w^2 + 42hw\\ \ \ \ \ \ \ \ \ \ \ \ = 26w^2 + 378w^{-1}\\ \text{Now, let's differentiate it for optimization.}\\ \ \ \ C'(w) = 52w - 378w^{-2}\\ \text{Let's determine the optimal width, }w^*.\\ \ \ \ C'(w*) = 52w^* - 378w^{*-2}\\ \ \ \ 0 = 52w^* - 378w^{*-2}\\ \ \ \ 378w^{*-2} = 52w^*\\ \ \ \ 378 = 52w^{*3}\\ \ \ \ w*^3 = \frac{378}{52}\\ \ \ \ w* = [\frac{378}{52}]^{\frac13}\\ \text{Now we know the optimal width, so let's calculate it's cost.}\\ \\C(w*) = 26[\frac{378}{52}]^\frac23 + 378[\frac{378}{52}]^{-\frac{1}3}\\ \ \ \ \ \ \ \ \ \ \ \simeq292.70 \]

OpenStudy (anonymous):

thanks guys!!!

OpenStudy (anonymous):

Who wants a medal? And who want me to be his fan?

ganeshie8 (ganeshie8):

lol you're my fan already n u gave me many medals before... :)

OpenStudy (anonymous):

I think whoever wants to manually simplify the cost at the end deserves as many medals as possible... :-)

OpenStudy (anonymous):

Oh K OB thanks ganeshie8! I'm ur fan in heart! :)

ganeshie8 (ganeshie8):

that word keeps me motivated me for next problem ;)

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