Thread: [SOLVED] Maximum error involved

A closed box to be found to have length 2ft, witdh 4ft and hight 3ft, where the measurement of each dimension is made with a maximum possible error of +-0.02ft. The top of the box is made from material that costs only 2 Dollars/$\displaystyle ft^2$. The material for the sides and bottom costs only 1.50 Dollars/$\displaystyle ft^2$. What is the maximum error involved in the computation of the cost of the box?

1) In that question, gonna find top,bottom and side surface areas with errors and calculate the cost?

2) Couldn't understand what it wants in question as answer. The difference between max and min value? ,, Only max value? or Only min value? ...

3) Just noticed that i dont know how to make multiplying with errors. So how can i calculate areas with errors?

p.s.--> I asked that question before but i just noticed that i didn't write the information about cost of sides and bottom. Now the question is complete.

Hello, Lafexlos!

A closed box to be found to have length 2ft, width 4ft and height 3ft.
The measurement of each dimension has a maximum possible error of ±0.02ft.

The top of the box is made from material that costs $2/ft².
The material for the sides and bottom costs $1.50/ft².
What is the maximum error involved in the computation of the cost of the box?

Let $\displaystyle L$ = length, $\displaystyle W$ = width, $\displaystyle H$ = height.

We are given: .$\displaystyle \begin{Bmatrix}L \:=\: 2 && dL \:=\: \pm0.02 \\ W \:=\: 4 && dW \:=\: \pm0.02 \\ H \:=\: 3 && dH \:=\: \pm0.02 \end{Bmatrix}$


The area of the top is $\displaystyle LW$ ft².
. . Its cost is: .$\displaystyle 2LW$ dollars.

The rest of the box has a total area of: $\displaystyle LW + 2LH + 2WH$ ft².
. . Its cost is: .$\displaystyle \tfrac{3}{2}(LW + 2LH + 2WH)$ dollars.

The total cost is: .$\displaystyle C \;=\;2LW + \tfrac{3}{2}(LW + 2LH + 2WH)$

. . We have: .$\displaystyle C \;=\;\tfrac{7}{2}LW + 3LH + 3WH $


Take differentials: .$\displaystyle dC \;=\;\tfrac{7}{2}(L\,dW + W\,dL) + 3(L\,dH + H\,dL) + 3(W\,dH + H\,dW)$

Substitute the given data:
. . $\displaystyle dC \;=\;\tfrac{7}{2}\bigg[2(0.02) + 4(0.02)\bigg] + 3\bigg[2(0.02) + 3(0.02)\bigg] + 3\bigg[4(0.02) + 3(0.02)\bigg] \;=\;1.14$


Therefore, the maximum error of the cost is: .$\displaystyle dC \:=\:\$1.14$

Arggh. Again that "error = differential" thing.
Ok. Thanks.
I think i won't forget from now on.