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:

$$\begin{gathered} a=60 \\ b=28 \\ V=(a-2c)(b-2c)c \\ V=(60-2c)(28-2c)c \\ V=4c^3-176c^2+1680c \\ {V}^{\mathrm{\prime }}=12c^2-352c+1680 \\ {V}^{\mathrm{\prime }}=0 \\ 12c^2-352c+1680=0 \\ 12c^2-352c+1680=0 \\ 12c^2-352c+1680=0 \\ p=12;q=-352;r=1680 \\ D=q^2-4pr=352^2-4\cdot 12\cdot 1680=43264 \\ D>0 \\ {c}_{1,2}={\displaystyle \frac{-q\pm \sqrt{D}}{2p}}={\displaystyle \frac{352\pm \sqrt{43264}}{24}} \\ {c}_{1,2}={\displaystyle \frac{352\pm 208}{24}} \\ {c}_{1,2}=14.66666667\pm 8.66666666667 \\ {c}_1=23.3333333333 \\ {c}_2=6 \\ \text{ Factored form of the equation: } \\ 12(c-23.3333333333)(c-6)=0 \\ c=c_1=23.3333=6 \\ {a}_1=a-2\cdot c_1=60-2\cdot 23.3333={\displaystyle \frac{40}{3}}\doteq 13.3333 \\ {b}_1=b-2\cdot c_1=28-2\cdot 23.3333=-{\displaystyle \frac{56}{3}}\doteq -18.6667 \\ {V}_1=a_1\cdot b_1\cdot c_1=13.3333\cdot (-18.6667)\cdot 23.3333\doteq -5807.4074 \\ {a}_2=a-2\cdot c_2=60-2\cdot 6=48 \\ {b}_2=b-2\cdot c_2=28-2\cdot 6=16 \\ {V}_2=a_2\cdot b_2\cdot c_2=48\cdot 16\cdot 6=4608 \\ {V}_2>V_1 \\ c=c_2=6=6\text{ cm} \end{gathered}$$

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