Respuesta :
Answer:
The dimensions that minimizes the cost are a= 6.604 (lenght of a side) and h=2.752 (height)
Step-by-step explanation:
Let's define some notation first:
h = represent the height of the box
a= length of a side on the bottom
We need to remember that we have on this case an open box, so then we don't have a top part.
Let's begin with the formula for the volume of the box, we have a total volume given of [tex]120m^3[/tex], and we know that the volume for a box is the product of the area and the height like this:
[tex]120 m^3 = a^2 h[/tex] (1)
From equation (1) we cna solve for the height and we got:
[tex]h=\frac{120}{a^2}[/tex] (2)
Now let's move to th surface area formula. The top bottom area is [tex]a^2[/tex] and the surface area for the four lateral sides is 4ah. Since we know that the cost for the bottom material is 10 $ per each square meter and for the sides is 12$ per square meter we can create a function cost like this:
[tex] Cost= Cost of bottom+ Cost of sides[/tex]
[tex]Cost = 10a^2 + (12)(4ah) = 10a^2 + 48ah[/tex] (3)
Now we can substitute into equation (3) the equation (2) like this:
[tex]Cost = 10a^2 + (12)(4ah) = 10a^2 + 48a\frac{120}{a^2}=10a^2 +\frac{5760}{a} [/tex] (4)
Now in order to find the minimum cost, we can take the first derivate for the expression (4) like this
[tex]\frac{d}{da} Cost = 20a -\frac{5760}{a^2}[/tex]
We can set this equal to 0 and solve for a and we got:
[tex]\frac{5760}{a^2}= 20a[/tex]
[tex]\frac{5760}{20}=a^3[/tex]
[tex]a=6.604[/tex]
And since we have a we can find h using the expression (2) like this:
[tex]h=\frac{120}{a^2}=\frac{120}{6.604^2}=2.752[/tex]
So then the dimensions that minimizes the cost are a= 6.604 (lenght of a side) and h=2.752 (height)
