A farmer can buy two types of plant food, mix A and mix B. Each cubic yard of mix A contains 35 pounds of phosphoric acid, 15
pounds of nitrogen, and 16 pounds of potash. Each cubic yard of mix B contains 14 pounds of phosphoric acid, 18 pounds of nitrogen, and 64 pounds of potash. The minimum monthly requirements are 840 pounds of phosphoric acid, 540 pounds of nitrogen, and 768 pounds of potash. If x is the number of cubic yards of mix A used and y is the number of cubic yards of mix B used, write a system of linear inequalities that indicates appropriate restraints on x and y. Find the set of feasible solutions graphically for the amounts of mix A and mix B that can be used.
1 answer:
Answer:
The answer is below
Step-by-step explanation:
Given that the minimum monthly requirements are 840 pounds of phosphoric acid, hence:
35 pounds of phosphoric acid mix A + 14 pounds of phosphoric acid mix B ≥ 840 pounds of phosphoric acid
35x + 14y ≥ 840
The minimum monthly requirements are 540 pounds of nitrogen, hence:
15 pounds of nitrogen mix A + 18 pounds of nitrogen mix B ≥ 540 pounds of nitrogen
15x + 18y ≥ 540
The minimum monthly requirements are 768 pounds of potash, hence:
16 pounds of potash mix A + 64 pounds of potash mix B ≥ 768 pounds of potash
16x + 64y ≥ 768
Also, x ≥ 0 and y ≥0.
Therefore the equations are:
35x + 14y ≥ 840
15x + 18y ≥ 540
16x + 64y ≥ 768
x ≥ 0 and y ≥ 0
Plotting with geogebra gives the graph below.
The point are (0,60), (48,0), (18,15), (32,4)
You might be interested in
The zeros of given function is – 5 and – 3
<u>Solution:</u>
We have to find the zeros of the function by rewriting the function in intercept form.
By using intercept form, we can put value of y as to obtain zeros of function
We know that, intercept form of above equation is
Taking “x” as common from first two terms and “3” as common from last two terms
x (x + 5) + 3(x + 5) = 0
(x + 5)(x + 3) = 0
Equating to 0 we get,
x + 5 = 0 or x + 3 = 0
x = - 5 or – 3
Hence, the zeroes of the given function are – 5 and – 3