Approximation by Increments

[yanarains]

A cubical box is to be constructed from three kinds of building materials. The materials used in the four sides of the box cost 4 cents per square inch, the material in the bottom costs 3 cents per square inch, and the material used for the lid costs 7 cents per square inch. Approximate the additional cost of all the building materials if the length of a side is increased from 19 inches to 19.5 inches (use the Approximation by Increments formula).


This is my work:
Cost=(area of base and top)(cost of square inch) + (area of sides)(cost of square inch)

C(x)= [2(x^2)](10)+[(4)(x)](4)
= [20x^2]+[16x]
=40x+16
Now with equation above I will find the cost if the length of the sides were to increase.
=[(40)(19)+16](.5)=388

i know that this is wrong. I need help setting up the equation correctly. Thank you

 

[Deleted member 4993]

A cubical box is to be constructed from three kinds of building materials. The materials used in the four sides of the box cost 4 cents per square inch, the material in the bottom costs 3 cents per square inch, and the material used for the lid costs 7 cents per square inch. Approximate the additional cost of all the building materials if the length of a side is increased from 19 inches to 19.5 inches (use the Approximation by Increments formula).


This is my work:
Cost=(area of base and top)(cost of square inch) + (area of sides)(cost of square inch)

C(x)= [2(x^2)](10)+[(4)(x)](4).........Why is the area of the sides only x

- it should be x^2 as well ....

The cost of lid and bottom should be 7*x^2 + 3*x^2 = 10*x^2

= [20x^2]+[16x]
=40x+16
Now with equation above I will find the cost if the length of the sides were to increase.
=[(40)(19)+16](.5)=388

i know that this is wrong. I need help setting up the equation correctly. Thank you