Science:Math Exam Resources/Courses/MATH152/April 2011/Question B 01 (a)

MATH152 April 2011

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Question B 01 (a)

Three identical fields of size one hectare are planted with different amounts of three kinds of wheat (types 1, 2 and 3). The first field is divided in three equal parts with one type of wheat planted on each part. It yields 12 tons of total harvest. The second field is divided in two equal parts with type 2 planted on one part and type 3 planted on the other. It yields 10 tons. The third field is divided in two equal parts with type 1 planted on one part and type 2 on the other. It yields 16 tons. Let $\mathbf {x} ={\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}$ be the vector of unknowns, where $x_{k}$ is the yield in tons per hectare of wheat type k. Describe the information above as as a linear system in the form $A\mathbf {x} =\mathbf {b}$ (write A and b with specific values).

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
How can we set up an equation for each field? How does that translate into a matrix?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We have measured the total output in tons. In a linear system sense, this would be the vector b. How do we relate this output to the input or unknowns, which in this case is the total amount of yield per hectare for each wheat type. This would be the vector x in Ax=b. If we consider a unit analysis we have to multiply x which is in tons per hectare by something that results in tons. Therefore, the entries of the matrix A must be hectares. We know that for the first field, a third of the area goes to each type of wheat. Since each field has a total area of 1 hectare, we know that the area taken up by each wheat in the first field is 1/3 hectare. Therefore we know that ${\frac {1}{3}}x_{1}+{\frac {1}{3}}x_{2}+{\frac {1}{3}}x_{3}=12$ since the total output is 12 tons. Similarly, for the second field we have that a half hectare is for type two and a half for type 3. Therefore we get a second equation $0x_{1}+{\frac {1}{2}}x_{2}+{\frac {1}{2}}x_{3}=10.$ Finally for the last field we have that a half hectare is for type 1 wheat and a half hectare for type two. Therefore, ${\frac {1}{2}}x_{1}+{\frac {1}{2}}x_{2}+0x_{3}=16.$ Notice we have three linear equations and thus we have a linear system. As mentioned above, we store the areas in the matrix A, the wheat type unknowns are our x and the resulting output is our b. Therefore we have ${\begin{aligned}A\mathbf {x} &=\mathbf {b} \\{\begin{bmatrix}{\frac {1}{3}}&{\frac {1}{3}}&{\frac {1}{3}}\\\\0&{\frac {1}{2}}&{\frac {1}{2}}\\\\{\frac {1}{2}}&{\frac {1}{2}}&0\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\\end{bmatrix}}&={\begin{bmatrix}12\\10\\16\end{bmatrix}}\end{aligned}}.$