Two cuboids

A cuboid has dimensions of 2 m × 3 m × 4 m. We increase the length of all edges by 50 cm.
1. By what percentage does the surface area of the cuboid increase compared to the original? Round the result to the nearest whole percent.
2. By what percentage does the volume of the cuboid increase compared to the original? Round the result to the nearest 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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