10.7 Volume and Surface Area

The surface area is $SA=2(10)(5)+2(5)(3)+2(10)(3)=190{\text{cm}}^2.$

The volume is $V=10(5)(3)=150{\text{cm}}^3.$

The area of the triangular base is $A_{base}=\frac{1}{2}(12)(6)=36{\text{in}}^2$. The perimeter of the base is $p=12+8.49+8.49=28.98\text{in}.$ Then, the surface area of the triangular prism is $SA=2(36)+28.98(10)=361.8{\text{in}}^2.$

The area of the triangular base is $A_{base}=\frac{1}{2}(6)(10.39)=31.17{\text{cm}}^2$. Then, the surface area is $SA=2(31.17)+36(10)=422.34{\text{cm}}^2.$

The volume formula is found by multiplying the area of the base by the height. We have that $V=B\cdot h=(31.17)(10)=311.7{\text{cm}}^3.$

The area of the long side measures $95\text{ft}\times 6.5\text{ft}=\mathrm{1,235}{\text{ft}}^2.$ Multiplying by 2 gives $\mathrm{2,470}{\text{ft}}^2.$ The front (minus the triangular area) measures $22\text{ft}\times 6.5\text{ft}=286{\text{ft}}^2.$ Multiplying by 2 gives $572{\text{ft}}^2.$ The floor measures $95\text{ft}\times 22\text{ft}=\mathrm{2,090}{\text{ft}}^2.$ Each triangular region measures $A=\frac{1}{2}(22)(5)=55{\text{ft}}^2.$ Multiplying by 2 gives $110{\text{ft}}^2.$ Finally, one side of the roof measures $12.1\text{ft}\times 95\text{ft}=\mathrm{1,149.5}{\text{ft}}^2.$ Multiplying by 2 gives $2299{\text{ft}}^2.$ Add them up and we have $SA=\mathrm{2,470}+572+\mathrm{2,090}+110+\mathrm{2,299}=\mathrm{7,541}{\text{ft}}^2.$

Step 1: We begin with the areas of the base and the top. The area of the circular base is

Step 2: The base and the top are congruent parallel planes. Therefore, the area for the base and the top is

Step 3: The area of the rectangular side is equal to the circumference of the base times the height:

Step 4: We add the area of the side to the areas of the base and the top to get the total surface area. We have

Step 5: The volume is equal to the area of the base times the height. Then,

The volume of the can of apple pie filling is $V=\pi {(4)}^2(10)=502.7{\text{cm}}^3.$ The volume of the pan is $V=\pi {(11)}^2(3)=1140.4{\text{cm}}^3.$ To find the number of cans of apple pie filling, we divide the volume of the pan by the volume of a can of apple pie filling. Thus, $1140÷502.7=\mathrm{2.3.}$ We will need 2.3 cans of apple pie filling to fill the pan.

So, how would we start? Let’s look at this on a smaller scale. Say that you have 30 inches of string and you experiment with different shapes. The rectangle in Figure 10.135 measures 12 in long by 3 in wide. We have a perimeter of $P=2(12)+2(3)=30\text{in}$ and the area calculates as $A=3(12)=36{\text{in}}^2.$ The rectangle in Figure 10.135, measures 8 in long and 7 in wide. The perimeter is $P=2(8)+2(7)=30\text{in}$ and the area is $A=8(7)=56{\text{in}}^2.$ In Figure 10.135, the square measures 7.5 in on each side. The perimeter is then $P=4(7.5)=30\text{in}$ and the area is $A=7.5(7.5)=56.25{\text{in}}^2.$ If you want a circular corral as in Figure 10.135, we would consider a circumference of $30=2\pi r,$ which gives a radius of $30÷2\pi =4.77\text{in}$ and an area of $A=\pi {(4.77)}^2=71.5{\text{in}}^2.$

Figure 10.135

We see that the circular shape gives the maximum area relative to a circumference of 30 in. So, a circular corral with a circumference of 150 meters and a radius of 23.87 meters gives a maximum area of $\mathrm{1,790.5}{\text{m}}^2.$

To choose the optimal shape for this container, you can start by experimenting with different sizes of boxes that will hold 3 cubic meters. It turns out that, similar to the maximum rectangular area example where a square gives the maximum area, a cube gives the maximum volume and the minimum surface area.

As all six sides are the same, we can use a simplified volume formula:

$V=s^3,$

where $s$ is the length of a side. Then, to find the length of a side, we take the cube root of the volume.

We have

$\begin{array}{ccc}\hfill 3& \hfill =\hfill & s^3\hfill \\ \hfill \sqrt[3]{3}& \hfill =\hfill & s\hfill \\ \hfill & \hfill =\hfill & 1.4422\text{m}\hfill \end{array}$

The surface area is equal to the sum of the areas of the six sides. The area of one side is $A={(1.4422)}^2=2.08{\text{m}}^2.$ So, the surface area of the crate is $SA=6(2.08)=12.5{\text{m}}^2$. At $15 a square meter, the cost comes to $12.5(\text{\$}15)=\text{\$}\mathrm{187.50.}$ Checking the volume, we have $V={(1.4422)}^3=2.99{\text{m}}^3$.