Question

A rectangular container with open top is required to have a volume of 24 cubic meters. Also, one side of the rectangular base is required to be 4 meters long. If material for the base costs $8 per square meter, and material for the side’s costs $2 per square meter, find the dimensions of the container so that the cost of material to make it will be a minimum.Step 1:a) make a diagramb) what formulas will be used in the problem?c) figure out what the constraint isd) what do you want to maximize or minimize?Step 2:a) write down formulas b) substitute in to formula you want to maximize or minimize c) find the critical points d) test the critical points e) answer the question

Answer 1

Given

A rectangular container with open top is required to have a volume of 24 cubic meters.

Also, one side of the rectangular base is required to be 4 meters long.

If material for the base costs $8 per square meter, and material for the side’s costs $2 per square meter.

To find the dimensions of the container so that the cost of material to make it will be a minimum.

Explanation:

a) It is given that,

b) The volume of the rectangular box is,

$V=l\cdot b\cdot h\text{ \_\_\_\_\_\_\lparen1\rparen}$

c) The constraint is,

$\begin{gathered} (1)\Rightarrow24=4bh \\ \Rightarrow bh=\frac{24}{4} \\ \Rightarrow bh=6cm^2 \end{gathered}$

d) To minimize the cost.

Step 2:

a) The formula is,

$Cost=Area\cdot(price\text{ }per\text{ }m^2)$

That implies, the cost for the materials for base is,

$\begin{gathered} Cost=(l\times b)\cdot(8) \\ C_b=4b\cdot(8) \\ =32b \end{gathered}$

Also, the cost for the materials for other sides is,

$\begin{gathered} C_R=(b\times h)\cdot(2) \\ =2bh \\ C_L=2bh \\ C_F=2lh \\ =2\times4h \\ =8h \\ C_B=8h \end{gathered}$

Hence, the total cost is,

$\begin{gathered} Total\text{ }Cost=C_b+C_R+C_L+C_F+C_B \\ C=32b+4bh+16h \end{gathered}$

b) Substitute bh=6cm^2 and h=(6/b) in C.

Then,

$\begin{gathered} C=32b+4\times6+16\times\frac{6}{b} \\ C=32b+24+\frac{96}{b} \\ \Rightarrow C=32b+96b^{-1}+24 \end{gathered}$

c) To find the critical points, find the derivative of C.

That implies,

$\begin{gathered} C^{\prime}=32+96(-1\times b^{-2}) \\ C^{\prime}=32-\frac{96}{b^2} \end{gathered}$

Also,

$\begin{gathered} C^{\prime\prime}=-96(-\frac{2}{b^3}) \\ =\frac{192}{b^3}>0 \end{gathered}$

Hence, the cost is minimum.

Now, set C'=0.

Then,

$\begin{gathered} 32-\frac{96}{b^2}=0 \\ \frac{96}{b^2}=32 \\ b^2=\frac{96}{32} \\ b^2=3 \\ b=\pm\sqrt{3} \end{gathered}$

Hence, the critical points are,

$\sqrt{3}\text{ }or\text{ }-\sqrt{3}$

d) To test the critical points, put the value of b in C'.

That implies,

$\begin{gathered} C^{\prime}=32-\frac{96}{b^2} \\ When\text{ }b=\sqrt{3}, \\ C^{\prime}=32-\frac{96}{(\sqrt{3})^2} \\ =32-\frac{96}{3} \\ =32-32 \\ =0 \\ ll^{rly},\text{ for }b=-\sqrt{3} \\ C^{\prime}=32-\frac{96}{(-\sqrt{3})^2} \\ =32-\frac{96}{3} \\ =32-32 \\ =0 \end{gathered}$

Hence, the critical points are tested.

e) Therefore, the dimensions of the container so that the cost is minimum is,

$\begin{gathered} l=4m,b=\sqrt{3}m,h=\frac{6}{\sqrt{3}} \\ l=4m,b=\sqrt{3}m,h=2\sqrt{3}m \end{gathered}$