how to do this??? (3 Viewers)

[jiffs]

i've spent like an hour on this

 

[C_master]

irrelevant: ignore what I voted above I was just checking to see peoples votes but theres none

 

[jiffs]

irrelevant: ignore what I voted above I was just checking to see peoples votes but theres none

lmao

 

[jiffs]

@carrotsss @tywebb @Trebla @synthesisFR @funnytomato @Drongoski
pls helpppppppppppppppppppp

 

[HazzRat]

we're given
(1) x + y ≥ a
(2) x − y ≤ −1
(3) the minimum value of z = x + ay is 7

As x ≤ y − 1, x ≥ a − y
So a − y ≤ x ≤ y − 1

Then sub different x values in, ensuring the minimum value is 7. You'd get a = -5 or 3.

 

[jiffs]

we're given
(1) x + y ≥ a
(2) x − y ≤ −1
(3) the minimum value of z = x + ay is 7

As x ≤ y − 1, x ≥ a − y
So a − y ≤ x ≤ y − 1

Then sub different x values in, ensuring the minimum value is 7. You'd get a = -5 or 3.

ur answer is wrong

 

[HazzRat]

ur answer is wrong

oh lmao. what's the correct one?

 

[jiffs]

oh lmao. what's the correct one?

b

 

[jiffs]

oh lmao. what's the correct one?

but how did u get ur answer tho?

 

[jiffs]

BUMP

 

[funnytomato]

There may be a better way than this but I think you could use graph to see it:

(1) x + y ≥ a
(2) x − y ≤ −1
Hence we have
y ≥ a - x
And y ≥ x + 1
Hence y would be in the region about either line (above a-x to the left of intersection and above x+1 to the right)

a) When a<0, there is no minimum to x + ay since you can keep the same x but make y larger and larger

b) When a>0, consider the minimum of x + ay : for the same x the value of y has to be minimised and hence it would be on the line a-x or x+1

b1) if 0<a<1, on the branch y=a-x, x+ay becomes x+a(a-x) = (1-a)x + a^2, again this has no minimum as 1-a>0 and as we keep decreasing x the sum x+ay keeps decreasing

b2) if a >1, then on the y=a-x branch (1-a)x + a^2 would be minimised when x largest as 1-a<0
Also on the y=x+1 branch x+ay becomes x(1+a)+a which is minimised when x is minimised since 1+a>0
Therefore the minimum can only be obtained in scenario b2 and that is exactly at the intersection of the 2 branches (you can then solve for that in terms of a and then equate to 7 you would get the answer noting a>1)