6 cm x cm 60 square cm
1 answer:
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Step-by-step explanation:
We have to find the point in the feasible region which maximizes the objective function. To find that point first we need to graph the given inequalities to find the feasible region.
Steps to graph 3x + y ≤ 12:
First we graph 3x + y = 12 then shade the graph for ≤.
plug any value of x say x=0 and x=2 into 3x + y = 12 to find points.
plug x=0
3x + y = 12
3(0) + y = 12
0 + y = 12
y = 12
Hence first point is (0,12)
Similarly plugging x=2 will give y=6
Hence second point is (2,6)
Now graph both points and joint them by a straight line.
test for shading.
plug any test point which is not on the graph of line like (0,0) into original inequality 3x + y ≤ 12:
3(0) + (0) ≤ 12
0 + 0 ≤ 12
0 ≤ 12
Which is true so shading will be in the direction of test point (0,0)
We can repeat same procedure to graph other inequalities.
From graph we see that ABCD is feasible region whose corner points will result into maximum or minimu for objective function.
Since objective function is not given in the question so i will explain the process.
To find the maximum value of objective function we plug each corner point of feasible region into objective function. Whichever point gives maximum value will be the answer
Answer:
Match each of the trigonometric expressions below with theequivalent non-trigonometric function from the following list.Enter the appropiate letter(A,B, C, D or E)in each blank
A . tan(arcsin(x/8))
B . cos (arsin (x/8))
C. (1/2)sin (2arcsin (x/8))
D . sin ( arctan (x/8))
E. cos (arctan (x/8))
These are the spaces to fill out :
.. ..........x/64 (sqrt(64-x^2))
.............x/sqrt(64+x^2)
.............sqrt(64-x^2)/8
..............x/sqrt(64-x^2)
..............8/sqrt(64+x^2)
A. ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)
B . cos (arsin (x/8)) ....sqrt(64-x^2)/8
Step-by-step explanation:
To solve this we have to find the missing sides to each of the triange discribed in prenthesis thus
A we have the sides of the triangle given by x, 8 and or
thus tan(arcsin(x/8)) = =
Therefore ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)
B
Here we have cos = adjacent/hypotenuse where adjacent side is and hypothenuse = 8 we have
/8
B . cos (arsin (x/8)) ....sqrt(64-x^2)/8
