Let the points (211,α) lie on or inside the triangle with sides x+y=11, x+2y=16, and 2x+3y=29. Then the product of the smallest and the largest values of α is equal to:
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Step-by-Step Solution
Step 1: Find the vertices of the triangle
First, we need to find the vertices of the triangle formed by the given lines. We will find the intersection points of each pair of lines. Let the lines be L1:x+y=11, L2:x+2y=16, and L3:2x+3y=29.
Step 2: Calculate intersection points
Solving L1 and L2: Subtracting L1 from L2 gives y=5. Substituting into L1 gives x=6. So, vertex A=(6,5).
Solving L2 and L3: Multiply L2 by 2 to get 2x+4y=32. Subtract L3 from this to get y=3. Substituting into L2 gives x+2(3)=16⟹x=10. So, vertex B=(10,3).
Solving L3 and L1: Multiply L1 by 2 to get 2x+2y=22. Subtract this from L3 to get y=7. Substituting into L1 gives x+7=11⟹x=4. So, vertex C=(4,7).
Wait, let's recheck the intersection of L2 and L3.
L2:x+2y=16⟹x=16−2y
Substitute into L3:2(16−2y)+3y=29⟹32−4y+3y=29⟹32−y=29⟹y=3.
Then x=16−2(3)=16−6=10. So B=(10,3).
Let's recheck the intersection of L1 and L2:
L1:x+y=11 L2:x+2y=16
Subtracting L1 from L2: (x+2y)−(x+y)=16−11⟹y=5.
Substitute y=5 into L1: x+5=11⟹x=6. So A=(6,5).
Let's recheck the intersection of L1 and L3:
L1:x+y=11⟹x=11−y L3:2x+3y=29
Substitute x=11−y into L3: 2(11−y)+3y=29⟹22−2y+3y=29⟹22+y=29⟹y=7.
Substitute y=7 into L1: x+7=11⟹x=4. So C=(4,7).
The vertices are A(6,5), B(10,3), and C(4,7).
Step 3: Substitute the point into the line equations
The point P(211,α) lies on or inside the triangle. This means that when we substitute the coordinates of P into the equations of the lines, the resulting values must satisfy the inequalities that define the interior of the triangle. To determine the correct inequality direction, we can test a point known to be inside the triangle, for example, the centroid. The centroid is (36+10+4,35+3+7)=(320,5).
For L1:x+y−11=0, Pc=320+5−11=320+15−33=32>0. So for points inside, x+y−11≥0.
For L2:x+2y−16=0, Pc=320+2(5)−16=320+10−16=320−6=320−18=32>0. So for points inside, x+2y−16≥0.
For L3:2x+3y−29=0, Pc=2(320)+3(5)−29=340+15−29=340−14=340−42=−32<0. So for points inside, 2x+3y−29≤0.
Now, substitute x=211 into these inequalities:
1. x+y−11≥0⟹211+α−11≥0⟹α−211≥0⟹α≥211
2. x+2y−16≥0⟹211+2α−16≥0⟹2α≥16−211⟹2α≥232−11⟹2α≥221⟹α≥421
3. 2x+3y−29≤0⟹2(211)+3α−29≤0⟹11+3α−29≤0⟹3α−18≤0⟹3α≤18⟹α≤6
Step 4: Determine the range of α
We have the following conditions for α:
α≥211=5.5 α≥421=5.25 α≤6
Combining these inequalities, we get:
α≥max(5.5,5.25)⟹α≥5.5
So, the range for α is 5.5≤α≤6.
Step 5: Calculate the product of smallest and largest values
The smallest value of α is 211 and the largest value of α is 6. The product of these values is 211×6=11×3=33.