Paper box

Hard rectangular paper has dimensions of 60 cm and 28 cm. The corners are cut off equal squares and the residue was bent to form an open box. How long must be side of the squares to be the largest volume of the box?

Correct result:

c =  6 cm

Solution:

a=60 b=28 V=(a−2c)(b−2c)c V=(60−2c)(28−2c)c V=4c3−176c2+1680c V′=12c2−352c+1680 V′=0 12c2−352c+1680=0  12c2−352c+1680=0 12c2−352c+1680=0  p=12;q=−352;r=1680 D=q2−4pr=3522−4⋅12⋅1680=43264 D>0  c1,2=−q±D2p=352±4326424 c1,2=352±20824 c1,2=14.66666667±8.66666666667 c1=23.3333333333 c2=6   Factored form of the equation:  12(c−23.3333333333)(c−6)=0 c=c1=23.3333=6 a1=a−2⋅ c1=60−2⋅ 23.3333≐403≐13.3333 b1=b−2⋅ c1=28−2⋅ 23.3333≐−563≐−18.6667 V1=a1⋅ b1⋅ c1=13.3333⋅ (−18.6667)⋅ 23.3333≐−5807.4074 a2=a−2⋅ c2=60−2⋅ 6=48 b2=b−2⋅ c2=28−2⋅ 6=16 V2=a2⋅ b2⋅ c2=48⋅ 16⋅ 6=4608 V2>V1 c=c2=6 cma=60 \ \\ b=28 \ \\ V=(a-2c)(b-2c)c \ \\ V=(60-2c)(28-2c)c \ \\ V=4c^3-176c^2+1680c \ \\ V'=12c^2- 352c +1680 \ \\ V'=0 \ \\ 12c^2- 352c +1680=0 \ \\ \ \\ 12c^2- 352c +1680=0 \ \\ 12c^2 -352c +1680=0 \ \\ \ \\ p=12; q=-352; r=1680 \ \\ D=q^2 - 4pr=352^2 - 4\cdot 12 \cdot 1680=43264 \ \\ D>0 \ \\ \ \\ c_{1,2}=\dfrac{ -q \pm \sqrt{ D } }{ 2p }=\dfrac{ 352 \pm \sqrt{ 43264 } }{ 24 } \ \\ c_{1,2}=\dfrac{ 352 \pm 208 }{ 24 } \ \\ c_{1,2}=14.66666667 \pm 8.66666666667 \ \\ c_{1}=23.3333333333 \ \\ c_{2}=6 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 12 (c -23.3333333333) (c -6)=0 \ \\ c=c_{1}=23.3333=6 \ \\ a_{1}=a-2 \cdot \ c_{1}=60-2 \cdot \ 23.3333 \doteq \dfrac{ 40 }{ 3 } \doteq 13.3333 \ \\ b_{1}=b-2 \cdot \ c_{1}=28-2 \cdot \ 23.3333 \doteq - \dfrac{ 56 }{ 3 } \doteq -18.6667 \ \\ V_{1}=a_{1} \cdot \ b_{1} \cdot \ c_{1}=13.3333 \cdot \ (-18.6667) \cdot \ 23.3333 \doteq -5807.4074 \ \\ a_{2}=a-2 \cdot \ c_{2}=60-2 \cdot \ 6=48 \ \\ b_{2}=b-2 \cdot \ c_{2}=28-2 \cdot \ 6=16 \ \\ V_{2}=a_{2} \cdot \ b_{2} \cdot \ c_{2}=48 \cdot \ 16 \cdot \ 6=4608 \ \\ V_{2} > V_{1} \ \\ c=c_{2}=6 \ \text{cm}

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