the same question came up a few days ago, were you the same person?
http://www.jiskha.com/display.cgi?id=1310104839
How many radios of each type can be constructed if the total number of transistors (P's, Q's and R's) available are 2200, 3400 and 1400 respectively and all transistors must be used?
http://www.jiskha.com/display.cgi?id=1310104839
I am stuck with this question, and I don't know how to solve it. Im getting x as -5 for some reason :S
Let's assume x, y, and z are the number of deluxe, standard, and economy radios produced, respectively.
From the given information, we know that:
For the deluxe radio: 2P + 7Q + 1R = 2200
For the standard radio: 2P + 3Q + 1R = 3400
For the economy radio: 1P + 2Q + 2R = 1400
Now, we can solve this system of equations.
First, let's solve for P:
Multiply the first equation by 2:
4P + 14Q + 2R = 2 * 2200
Rearrange the second equation:
4P + 6Q + 2R = 6800
Subtract the rearranged equation from the first equation:
(4P + 14Q + 2R) - (4P + 6Q + 2R) = 2200 - 6800
8Q = -4600
Q = -4600 / 8
Q = -575
Since we can't have a negative number of transistors, this means there is no solution to the system of equations given.
Therefore, we cannot construct any radios of each type using the given number of transistors.