June 14th, 2017, 03:46 AM  #1 
Newbie Joined: Feb 2016 From: west midlands Posts: 2 Thanks: 0  Differentiation
Problem 19. An open rectangular box with square ends is fitted with an overlapping lid which covers the top and the front face. Determine the maximum volume of the box if 11m² of metal are used in its construction. Would the answer be 4/5m³? Last edited by skipjack; June 14th, 2017 at 07:31 AM. 
June 14th, 2017, 04:02 AM  #2 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित 
Lid's geometry is not clear.

June 14th, 2017, 04:07 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
What does "covers the top and front face" mean? I imagine a piece bent at right angles so one half sits on top of the open top and the other bends down in front of one face. Is that correct? If so, then let the base be x by x meter and the sides y meters high. The volume is [math]x^2y[/mayth] cubic meters. The base has area $\displaystyle x^2$ square meters and there are four sides each of area $\displaystyle xy$ square meters. The "lid that covers the top and front face" has area $\displaystyle x^2+ xy$ square meters so a total area of $\displaystyle x^2+ 4xy+ x^2+ xy= 2x^2+ 5xy$ square meters is required. We want to maximize $\displaystyle x^2y$ with the constraint $\displaystyle 2x^2+ 5xy= 11$. We can write the constraint as $\displaystyle 5xy= 11 2x^2$ and then solve for y: $\displaystyle y= \frac{11 2x^2}{5x}$. Putting that into the "object" function, we want to maximize $\displaystyle x^2\frac{11 2x^2}{5x}= (11/5)x (2/5)x^3$. Last edited by skipjack; June 14th, 2017 at 07:37 AM. 
June 14th, 2017, 08:47 AM  #4 
Newbie Joined: Dec 2016 From: Austin Posts: 16 Thanks: 1  Solution
You'll need to create an equation for the surface area and volume of the box. Label the square ends with dimensions of $\displaystyle x$ and the longer front side $\displaystyle y$. Find the primary volume equation that we're attempting to optimize: $\displaystyle V=x^{2}y$ Since you have two square ends and a lid that covers the top and front face, we can create a secondary equation for the surface area: $\displaystyle A = 2x^{2}+5xy$ $\displaystyle 2x^{2}+5xy=11$ Notice that the volume equation has two variables. We can get this in terms of one variable by solving the secondary equation for $\displaystyle y$: $\displaystyle 2x^{2}+5xy=11$ $\displaystyle 5xy=112x^{2}$ $\displaystyle y=\frac{ 112x^{2} }{5x}$ Substitute this into the primary equation in order to get it in terms of one variable: $\displaystyle V=x^{2}y$ $\displaystyle V=x^{2} \left( \frac{ 112x^{2} }{5x} \right )$ $\displaystyle V=\frac{11}{5}x\frac{2}{5}x^{3}$ Differentiate, set equal to zero, and solve: $\displaystyle \frac{dV}{dx} = \frac{11}{5}\frac{6}{5}x^{2}$ $\displaystyle \frac{11}{5}\frac{6}{5}x^{2}=0$ $\displaystyle \frac{6}{5}x^{2}=\frac{11}{5}$ $\displaystyle 6x^{2}=11$ $\displaystyle x^{2}=\frac{11}{6}$ $\displaystyle x=\sqrt{ \frac{11}{6} }$ $\displaystyle x=\frac{ \sqrt{66} }{6}$ Now substitute this into the secondary equation to solve for $\displaystyle y$: $\displaystyle y=\frac{ 112x^{2} }{5x}$ $\displaystyle y = \frac{112 \left( \frac{11}{6} \right ) }{ 5 \left( \frac{ \sqrt{66} }{6} \right ) }$ $\displaystyle y = \frac{ \frac{22}{3} }{ \frac{5 \sqrt{66}}{6}}$ $\displaystyle y = \frac{22}{3} \left ( \frac{6}{5 \sqrt{66} } \right )$ $\displaystyle y = \frac{44}{5 \sqrt{66}}$ $\displaystyle y = \frac{44 \sqrt{66}}{ 5(66)}$ $\displaystyle y = \frac{2 \sqrt{66}}{15}$ Now that you have the dimensions, substitute them back into the volume equation, to find the maximum volume: $\displaystyle V=x^{2}y$ $\displaystyle V = \left ( \frac{11}{6} \right ) \left ( \frac{2 \sqrt{66} }{15} \right )$ $\displaystyle V = \frac{ 11 \sqrt{66} }{45}$ $\displaystyle V \approx 1.99 \text{ m}^{3}$ Hope that helps! Please let me know if you find an error anywhere, and I'll correct it. 

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calculus, differentiation, integration, problem solving, volume 
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