A rectangular box with a square bottom and closed top is to be made from two materials. The material for the side costs $1.50 per square foot and the material for the top and bottom costs $3.00 per square foot. If you are willing to spend $15 on the box, what is the largest volume it can contain? Justify your answer completely using calculus.

If anyone can give step-by-step instructions on how to do this problem, I would really appreciate it. Thanks in advance.

gibsonj338 Jan 9, 2019

#1**+4 **

\(\text{the wording is kind of odd. Is the bottom open? Sort of a weird way to orient the problem if so}\\ \text{I'm going to assume both the top and bottom are closed}\\ \text{we've got the two different types of materials so let's compute the two areas separately}\\ \text{assume the box has square side length }s \text{ and height }h\\ area_{side}=4sh\\ area_{tb} = 2s^2\\ vol = s^2h\)

\(\text{we can convert areas into cost using the materials prices}\\ C = 4sh\cdot (1.5) + 2s^2 \cdot (3) = 6sh + 6s^2\)

\(\text{I"m going to assume you haven't been taught Lagrange multipliers yet}\\ \text{so we have to use the cost function to write the volume in a single variable}\\ 15 = 6sh + 6s^2\\ h = \dfrac{15-6s^2}{6s}\\ vol = s^2 h = s^2\dfrac{15-6s^2}{6s} = \dfrac{15s-6s^3}{6}\)

\(\text{Now to find the extrema of the volume we set it's derivative equal to 0}\\ \dfrac{dvol}{ds} = \dfrac{15-18s^2}{6} =\dfrac{ 5-6s^2}{2} = 0\\ 5 = 6s^2\\ s = \sqrt{\dfrac 5 6}~ft \text{ (we can ignore the negative solution)}\)

\(vol = \dfrac 1 6 \left(15\sqrt{\dfrac 5 6} -6 \left(\dfrac 5 6\right)^{3/2}\right) = \dfrac 5 3 \sqrt{\dfrac 5 6}~ft^3\)

.Rom Jan 10, 2019

#3**+2 **

The top is closed per the question....but it does not say if it is the same material as the bottom.....just states BOTTOM is 3/sqft.....

I'm glad you got an answer! I got stuck when I tried top made of same material as sides and only bottom at $3/sqft. ~EP

ElectricPavlov
Jan 10, 2019

#2**+2 **

\(\text{What's that? Lagrange multipliers?}\\ \text{The method of Lagrange multipliers involves solving}\\ \nabla (f(s,h) - \lambda (c(s,h)-C))\\ \text{In this problem }\\ f(s,h) = vol(s,h) = s^2 h\\ c(s,h)=6sh+6s^2 (\text{see the previous answer})\\ C=15\)

\(\text{our Lagrange equation becomes}\\ \nabla \left(s^2 h - \lambda (6sh-6s^2 - 15)\right)= (0,0,0)\\ \text{this leads to the system of equations }\\ 2sh - \lambda(6h - 12s)=0\\ s^2 - 6s\lambda = 0\\ 6sh-6s^2-15 = 0\\\)

\(\text{and these are solved for }s,~h,~\lambda\\ \text{This seems like a lot more work than the previous answer. Why do it?}\\ \text{In most problems you can't cleanly solve for one in terms of the other and substitute like we did}\)

.Rom Jan 10, 2019

#4**+1 **

Let the volume be

V = x^2*h where x is the side of the base and h is the height (1)

And.....using the total cost for the surface area....we know that

Total cost for top/ bottom + Total cost for sides = 15

2x^2(3) + 4xh(1.5) = 15

15 - 6x^2 = 6xh

[ 15 - 6x^2] / [6x] = h (2)

Put (2) into (1) and we have that

V = x^2 [ 15 - 6x^2] / [ 6x]

V = x [ 15 - 6x^2] / 6

V = (1/6) (15x - 6x^3 ) take the derivative and set to 0

(1/6) (15 - 18x^2) = 0

18x^2 = 15

x^2 = 15/18 = 5/6

x = √[5/6]

So.....the volume of the box is : (1/6)√[5/6] (15 -6 (5/6) ) = (1/6)√[5/6] (10) = (5/3)√[5/6] ft^3

CPhill Jan 10, 2019