Two cuboids

The cuboid has dimensions of 2m 3m 4m We increase the length of all edges by 50 cm.
1. Calculate in % how much the surface area of the cuboid increases compared to the surface area of the original cuboid - round the result to whole percent
2. Calculate in % how much the volume of the cuboid increases compared to the volume of the original cuboid - round the result to whole percent

Final Answer:

p₁ =  38 %
p₂ =  64 %

Step-by-step explanation:

$$\begin{gathered} a=2\text{m} \\ b=3\text{m} \\ c=4\text{m} \\ \Delta =50\text{cm}\to \text{m}=50:100\text{m}=0.5\text{m} \\ A=a+\Delta =2+0.5=\frac{5}{2}=2.5\text{m} \\ B=b+\Delta =3+0.5=\frac{7}{2}=3.5\text{m} \\ C=c+\Delta =4+0.5=\frac{9}{2}=4.5\text{m} \\ {S}_1=2\cdot (a\cdot b+b\cdot c+c\cdot a)=2\cdot (2\cdot 3+3\cdot 4+4\cdot 2)=52{\text{m}}^2 \\ {S}_2=2\cdot (A\cdot B+B\cdot C+C\cdot A)=2\cdot (2.5\cdot 3.5+3.5\cdot 4.5+4.5\cdot 2.5)=\frac{143}{2}=71.5{\text{m}}^2 \\ {p}_1=100\cdot \frac{S_2-S_1}{S_1}=100\cdot \frac{71.5-52}{52}=38\% \end{gathered}$$

$$\begin{gathered} V_1=a\cdot b\cdot c=2\cdot 3\cdot 4=24{\text{m}}^3 \\ {V}_2=A\cdot B\cdot C=2.5\cdot 3.5\cdot 4.5=\frac{315}{8}=39.375{\text{m}}^3 \\ {p}_2=100\cdot \frac{V_2-V_1}{V_1}=100\cdot \frac{39.375-24}{24}=64\% \end{gathered}$$

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