OpenStudy (anonymous):

A clothing manufacturer has a $7 million budget for advertising for the coming year. There are four categories for the expenditure: television, radio, magazines and direct mail.$4 million should be spent on television and radio combined. The total spent on radio and magazines should equal the amount spent on television. All the advertising should be spaced evenly throughout the year, except for direct mail, which should be evenly spaced in the four months June through Sept. Only \$3 million should be spent in the first six months of the year. Decide how much should be spent in each category.

5 years ago
OpenStudy (anonymous):

t + r + m + d = 7. t + r = 4. -t + r + m = 0. So far.

5 years ago
OpenStudy (anonymous):

(1/2)t + (1/2)r + (1/2)m + (1/4)d = 3. There, that is the last linear equation. You now have 4 equations in 4 variables, so you should be able to solve it from here if you are somewhat familiar with linear algebra. If you still need help, then just ask a question. (1/4)d is from June which falls in the first half of the year. Making sense to you?

5 years ago
OpenStudy (anonymous):

yes this helps alot thank u

5 years ago
OpenStudy (anonymous):

I saw that your post was 2 days old? No one helped you before this time?

5 years ago
OpenStudy (anonymous):

no

5 years ago
OpenStudy (anonymous):

By the way, I set up a matrix for the coefficients, and the determinant is not = 0, so the matrix is invertible, which means that the 4 equations wil DEFINITELY give the answer. If you have any problems with this further down the line of your solving, just ask.

5 years ago
OpenStudy (anonymous):

thank u im working on the matrix right now to get an asnwer this helped alot

5 years ago
OpenStudy (anonymous):

Maybe you're done already with the final answer. If not, I went through the effort to get the final answer and plugged the answers back into the matrix, and it all works out nicely from beginning to end. Just in case you're stuck. The method I used to get the final solution was to get the inverse of the original matrix.

5 years ago