Paper box

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

Correct answer:

c =  6 cm

Step-by-step explanation:

$$\begin{gathered} a=60 \\ b=28 \\ V=(a-2c)(b-2c)c \\ V=(60-2c)(28-2c)c \\ V=4c^3-176c^2+1680c \\ {V}^{\prime }=12c^2-352c+1680 \\ {V}^{\prime }=0 \\ 12c^2-352c+1680=0 \\ 12c^2-352c+1680=0 \\ p=12;q=-352;r=1680 \\ D=q^2-4pr=352^2-4\cdot 12\cdot 1680=43264 \\ D>0 \\ {c}_{1,2}=\frac{-q\pm \sqrt{D}}{2p}=\frac{352\pm \sqrt{43264}}{24} \\ {c}_{1,2}=\frac{352\pm 208}{24} \\ {c}_{1,2}=14.666667\pm 8.666667 \\ {c}_1=23.333333333 \\ {c}_2=6 \\ c=c_1=23.3333=6 \\ {a}_1=a-2\cdot c_1=60-2\cdot 23.3333=\frac{40}{3}\doteq 13.3333 \\ {b}_1=b-2\cdot c_1=28-2\cdot 23.3333=-\frac{56}{3}\doteq -18.6667 \\ {V}_1=a_1\cdot b_1\cdot c_1=13.3333\cdot (-18.6667)\cdot 23.3333\doteq -5807.4074 \\ {a}_2=a-2\cdot c_2=60-2\cdot 6=48 \\ {b}_2=b-2\cdot c_2=28-2\cdot 6=16 \\ {V}_2=a_2\cdot b_2\cdot c_2=48\cdot 16\cdot 6=4608 \\ {V}_2>V_1 \\ c=c_2=6=6\text{cm} \end{gathered}$$

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