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 the largest volume of the box be beside the squares?

Correct answer:

c =  6 cm

Step-by-step explanation:

$$\begin{gathered} a=60 \\ b=28 \\ V=(a-2c)(b-2c)c \\ V=(60-2c)(28-2c)c \\ V=4c^3-176c^2+1680c \\ {V}^{\prime }=12c^2-352c+1680 \\ {V}^{\prime }=0 \\ 12c^2-352c+1680=0 \\ 12c^2-352c+1680=0 \\ 12=2^2\cdot 3 \\ 352=2^5\cdot 11 \\ 1680=2^4\cdot 3\cdot 5\cdot 7 \\ \text{GCD}(12,352,1680)=2^2=4 \\ 3c^2-88c+420=0 \\ p=3;q=-88;r=420 \\ D=q^2-4pr=88^2-4\cdot 3\cdot 420=2704 \\ D>0 \\ {c}_{1,2}=\frac{-q\pm \sqrt{D}}{2p}=\frac{88\pm \sqrt{2704}}{6} \\ {c}_{1,2}=\frac{88\pm 52}{6} \\ {c}_{1,2}=14.666667\pm 8.666667 \\ {c}_1=23.333333333 \\ {c}_2=6 \\ c=c_1=23.3333=6 \\ {a}_1=a-2\cdot c_1=60-2\cdot 23.3333=\frac{40}{3}\doteq 13.3333 \\ {b}_1=b-2\cdot c_1=28-2\cdot 23.3333=-\frac{56}{3}\doteq -18.6667 \\ {V}_1=a_1\cdot b_1\cdot c_1=13.3333\cdot (-18.6667)\cdot 23.3333\doteq -5807.4074 \\ {a}_2=a-2\cdot c_2=60-2\cdot 6=48 \\ {b}_2=b-2\cdot c_2=28-2\cdot 6=16 \\ {V}_2=a_2\cdot b_2\cdot c_2=48\cdot 16\cdot 6=4608 \\ {V}_2>V_1 \\ c=c_2=6=6\text{cm} \end{gathered}$$

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