Asked by Dan
                An open-topped rectangular box is to be constructed from a 24 inch by 36 inch piece of cardboard by cutting
out squares of equal sides from the corners and then folding up the sides. What size squares should be cut out
of each of the corners in order to maximize the volume of the box and what is the maximum volume?
I tried this out by multiplying the width and the length together and got 896.
I then went ahead and plugged it into the formula for a rectangular prism with x as my variable => V(x)=x(896-2x)^2
Unfortunately, I realized that this would lead me to get a positive second derivative as an answer. I wouldn't get a max with that, rather, a min.
Not sure where I went wrong with this one.
            
        out squares of equal sides from the corners and then folding up the sides. What size squares should be cut out
of each of the corners in order to maximize the volume of the box and what is the maximum volume?
I tried this out by multiplying the width and the length together and got 896.
I then went ahead and plugged it into the formula for a rectangular prism with x as my variable => V(x)=x(896-2x)^2
Unfortunately, I realized that this would lead me to get a positive second derivative as an answer. I wouldn't get a max with that, rather, a min.
Not sure where I went wrong with this one.
Answers
                    Answered by
            Damon
            
    new length = 36 - 2x
new width = 24 - 2 x
so
V = (36-2x)(24-2x)x
= 864x - 120 x^2 + 4 x^3
I am glad you do derivatives because I do not feel like finding the vertex.
dV/dx = 864 -240 x + 12 x^2
0 = 72 - 20 x + x^2
x = [ 20 +/- sqrt(400 -288) ]/2
= [ 20 +/- 10.6 ] /2
= 15.3 or 4.7
4.7 works
0 when x = 120/8 = 15
    
new width = 24 - 2 x
so
V = (36-2x)(24-2x)x
= 864x - 120 x^2 + 4 x^3
I am glad you do derivatives because I do not feel like finding the vertex.
dV/dx = 864 -240 x + 12 x^2
0 = 72 - 20 x + x^2
x = [ 20 +/- sqrt(400 -288) ]/2
= [ 20 +/- 10.6 ] /2
= 15.3 or 4.7
4.7 works
0 when x = 120/8 = 15
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