2
$\begingroup$

Producing $x_1$ trucks and $x_2$ planes requires $x_1+50x_2$ tons of steel, $40x_1+1000x_2$ pounds of rubber, and $2x_1+50x_2$ months of labor. If the unit costs $y_1, y_2, y_3$ are \$700 per ton, \$3 per pound, and \$3000 per month, what are the values of one truck and one plane? Those are the components of $(A^T)y$.

I don't understand the role of the unit costs $y_1, y_2, y_3$. How do I set up the system for this problem? And why is it $(A^T)y$?

This problem is from the book Linear Algebra and Its Applications 4ed by Prof Gilbert Strang, page 65, chapter 1.6, problem 62.

$\endgroup$
3
  • 1
    $\begingroup$ Check out the following link to help format your question. meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Vincent Aug 29 '14 at 16:59
  • $\begingroup$ what's $A$? Please define all the terms. $\endgroup$ – Memming Aug 29 '14 at 17:19
  • $\begingroup$ This looked like a linear programming problem to me, rather than linear algebra... $\endgroup$ – Diya Aug 29 '14 at 17:32
2
$\begingroup$

Let $A=\begin{bmatrix}1&50\\40&1000\\2&50\end{bmatrix}$. If $x=\begin{bmatrix}x_1\\x_2\end{bmatrix}$, then $Ax$ gives the amount of resources (steel, rubber, and labor) to build $x_1$ trucks and $x_2$ planes;

and the first column of $A$ gives the resources required for 1 truck, while the second column gives the resources required for 1 plane.

If $y=\begin{bmatrix}700\\3\\3000\end{bmatrix}$, then $y$ gives the unit costs for steel, rubber, and labor; so the entries of

$A^{T}y=\begin{bmatrix}1&40&2\\50&1000&50\end{bmatrix}\begin{bmatrix}700\\3\\3000\end{bmatrix}=\begin{bmatrix}6,820\\188,000\end{bmatrix}$ give the cost to make one truck and one plane.

$\endgroup$
0
$\begingroup$

To calculate the costs for one truck, you have to set $x_1$ equal to 1. Then you have to multiply the coefficients for $x_1$ with the costs:

Costs for one truck: $1 \cdot 700+ 40 \cdot 3+2\cdot 3,000=6,820$

Costs for one plane: similar calculation.

The term for calculations can be written by matrices $\left( A^T \right) y$:

$ \begin{pmatrix}1 & 40 & 2 \\50 & 1000 & 50 \\\end{pmatrix}\times \begin{pmatrix} 700 \\ 3 \\ 3000 \end{pmatrix} $

$\endgroup$
0
$\begingroup$

This sort of thing always confused me until I started thinking of matrices not as things, but as functions.

An $m \times n$ matrix is a function from $R^n$ to $R^m$, from vectors with $n$ elements to vectors with $m$ elements. In this problem, we start out with the $3 \times 2$ matrix

$$\mathbf{A} = \begin{bmatrix}1 & 50\\ 40 & 1000\\ 2 & 50\end{bmatrix}$$

which, when we multiply it by a vector $x = \begin{bmatrix}\text{number of trucks}\\\text{number of planes}\end{bmatrix} = \begin{bmatrix}x_1\\x_2\end{bmatrix}$, yields:

$$\begin{bmatrix}1 & 50\\ 40 & 1000\\ 2 & 50\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}x_1 + 50x_2\\40x_1 + 1000x_2\\2x_1 + 50x_2\end{bmatrix} = \begin{bmatrix}\text{tons of steel}\\\text{pounds of rubber}\\\text{months of labor}\end{bmatrix}$$

So the matrix $\mathbf{A}$ acted like a function: it took in a vector of quantities in trucks-planes space and output a vector of quantities in steel-rubber-labor space.

(This is the reason why an $m \times n$ matrix times a $n \times 1$ matrix will give an $m \times 1$ matrix: we're translating from $R^n$ to $R^m$).

Okay, so as we've got it right now, $\mathbf{A}$ sends trucks-planes to steel-rubber-labor. What matrix will send steel-rubber-labor (now in \$, but that's just a change of units) to trucks-planes? $\mathbf{A}^T$.

$$\begin{bmatrix}1 & 40 & 2\\50 & 1000 & 50\end{bmatrix}\begin{bmatrix}\$700\\\$3\\\$3000\end{bmatrix} = \begin{bmatrix}\$700 + \$120 + \$6000\\\$35000 + \$30000 + \$150000\end{bmatrix} = \begin{bmatrix}\$6820\\\$188000\end{bmatrix}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.