Respuesta :
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 = [tex]\dfrac{350}{ac}[/tex]
replacing the value for c in the above equation (1); we have :
[tex]C = 8 ac + 14 a(\dfrac{350}{ac}) + 8c(\dfrac{350}{ac})[/tex]
[tex]C = 8 ac + \dfrac{4900}{c}+\dfrac{2800}{a}[/tex]
Differentiating C with respect to a and c; we have:
[tex]C_a = 8c - \dfrac{2800}{a^2}[/tex]
[tex]C_c = 8a - \dfrac{4900}{c^2}[/tex]
[tex]8c - \dfrac{2800}{a^2}=0[/tex] --- (2)
[tex]8a - \dfrac{4900}{c^2}=0[/tex] ---(3)
From (2)
[tex]8c =\dfrac{2800}{a^2}[/tex]
[tex]c =\dfrac{2800}{8a^2}[/tex] ----- (4)
From (3)
[tex]8a =\dfrac{4900}{c^2}[/tex]
[tex]a =\dfrac{4900}{8c^2}[/tex] -----(5)
Replacing the value of a in 5 into equation (4)
[tex]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[/tex]
From (5)
[tex]a =\dfrac{4900}{8c^2}[/tex] -----(5)
[tex]a =\dfrac{4900}{8* 10.234^2}[/tex]
a = 5.8481
Recall that :
b = [tex]\dfrac{350}{ac}[/tex]
b = [tex]\dfrac{350}{5.8481*10.234}[/tex]
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
The dimensions that will minimize the cost of constructing this box are: a = 5.8481 inches, b = 5.848 inches, and c = 10.234 inches and this can be determined by using the given data.
Given :
- An open-top rectangular box is being constructed to hold a volume of 350 inches cube.
- The base of the box is made from a material costing 8 cents/inch square.
- The front of the box must be decorated and will cost 10 cents/inch square.
- The remainder of the sides will cost 4 cents/inch square.
According to the given data the total cost is given by:
C = 8ac + 14ab + 8cb --- (1)
The volume of the rectangular box is (V = abc = 350 inch cube). So, the value of b is given by:
[tex]\rm b = \dfrac{350}{ac}[/tex]
Now, substitute the value of 'b' in the equation (1).
[tex]\rm C = 8ac + \dfrac{4900}{c}+\dfrac{2800}{a}[/tex]
First differentiating the above equation with respect to c.
[tex]\rm C_c = 8a-\dfrac{4900}{c^2}[/tex] --- (2)
Now, differentiating the above equation with respect to a.
[tex]\rm C_a = 8c-\dfrac{2800}{a^2}[/tex] --- (3)
Now, equate equation (2) and equation (3) to zero.
From equation (2):
[tex]\rm a=\dfrac{4900}{8c^2}[/tex] ----- (4)
From equation (3):
[tex]\rm c=\dfrac{2800}{8a^2}[/tex] ----- (5)
Now, from equations (4) and (5).
[tex]\rm c = \dfrac{2800}{8\left(\dfrac{4900}{8c^2}\right)^2}[/tex]
Now, simplifying the above expression in order to get the value of c.
c = 10.234
Now, put the value of 'c' in equation (5) in order to get the value of 'a'.
a = 5.8481
The value of 'b' is given by:
[tex]\rm b = \dfrac{350}{5.8481\times 10.234}[/tex]
b = 5.848
So, the dimensions that will minimize the cost of constructing this box are: a = 5.8481 inches, b = 5.848 inches, and c = 10.234 inches.
For more information, refer to the link given below:
https://brainly.com/question/19770987