Question

An open-top rectangular box is being constructed to hold a volume of 350 in3. The base of the box is made from a material costing 8 cents/in2. The front of the box must be decorated, and will cost 10 cents/in2. The remainder of the sides will cost 4 cents/in2. Find the dimensions that will minimize the cost of constructing this box.

Answer 1

Answer:

the dimensions that will minimize the cost of constructing the box is:

a = 5.8481  in ;   b = 5.848 in  ; c = 10.234 in

Step-by-step explanation:

From the information given :

Let a be the base if the rectangular box

b to be the height and c to be the  other side of the rectangular box.

Then ;

the area of the base is ac

area for the front of the box is ab

area for the remaining other sides   ab + 2cb

The base of the box is made from a material costing 8 ac

The front of the box must be decorated, and will cost 10 ab

The remainder of the sides will cost 4 (ab + 2cb)

Thus ; the total cost  C is:

C = 8 ac + 10 ab + 4(ab + 2cb)

C = 8 ac + 10 ab + 4ab + 8cb

C = 8 ac + 14 ab + 8cb   ---- (1)

However; the volume of the rectangular box is V = abc = 350 in³

If abc = 350

Then b = $\dfrac{350}{ac}$

replacing the value for c in the above equation (1); we have :

$C = 8 ac + 14 a(\dfrac{350}{ac}) + 8c(\dfrac{350}{ac})$

$C = 8 ac + \dfrac{4900}{c}+\dfrac{2800}{a}$

Differentiating C with respect to a and c; we have:

$C_a = 8c - \dfrac{2800}{a^2}$

$C_c = 8a - \dfrac{4900}{c^2}$

$8c - \dfrac{2800}{a^2}=0$ --- (2)

$8a - \dfrac{4900}{c^2}=0$   ---(3)

From (2)

$8c =\dfrac{2800}{a^2}$

$c =\dfrac{2800}{8a^2}$ ----- (4)

From (3)

$8a =\dfrac{4900}{c^2}$

$a =\dfrac{4900}{8c^2}$   -----(5)

Replacing the value of a in 5 into equation (4)

$c = \dfrac{2800}{8*(\dfrac{4900}{8c^2})^2} \\ \\ \\ c = \dfrac{2800}{\dfrac{8*24010000}{64c^4}} \\ \\ \\ c = \dfrac{2800}{\dfrac{24010000}{8c^4}} \\ \\ \\ c = \dfrac{2800*8c^4}{24010000} \\ \\ c = 0.000933c^4 \\ \\ \dfrac{c}{c^4}= 0.000933 \\ \\ \dfrac{1}{c^3} = 0.000933 \\ \\ \dfrac{1}{0.000933} = c^3 \\ \\ 1071.81 = c^3\\ \\ c= \sqrt[3]{1071.81} \\ \\ c = 10.234$

From (5)

$a =\dfrac{4900}{8c^2}$   -----(5)

$a =\dfrac{4900}{8* 10.234^2}$

a = 5.8481

Recall that :

b = $\dfrac{350}{ac}$

b = $\dfrac{350}{5.8481*10.234}$

b =5.848

Therefore ; the dimensions that will minimize the cost of constructing the box is:

a = 5.8481  in ;   b = 5.848 in  ; c = 10.234 in