A straight line L with negative slope passes through (8, 2) and cuts the positive axes at P and Q.
As L varies, the absolute minimum value of OP + OQ is (O is origin)
(A) 10 (B) 18
(C) 16 (D) 12
Answers
Answered by
54
let slope = - m , where m > 0.
let point A be (8, 2).
equation of L : y = - m x + c. It passes through A.
2 = - m * 8 + c => c = 2 + 8 m
OP = x intercept , ie., value of x when y = 0.
0 = - m * OP + c = - m * OP + 2 + 8 m
=> OP = 2 / m + 8
OQ = y intercept , ie., value of y when x = 0
OQ = c = 2 + 8 m
OP + OQ = 8 m + 10 + 2 / m
derivative of (OP + OQ) wrt m : 8 - 2 / m²
derivative = 0 when: m = +1/2 or -1/2. we take only the positive value.
minimum value of OP + OQ = 4 + 10 + 4 = 18
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simpler method:
The equation of L in the intercept form: x/OP + y/OQ = 1
We are given that OP and OQ are positive. As L passes through A (8, 2):
8/OP + 2/OQ = 1
8 OQ + 2 OP = OP * OQ
OP = 8 OQ / (OQ - 2)
The sum of intercepts : OP + OQ = 8 OQ / (OQ - 2) + OQ
Derivative of OP + OQ wrt OQ: [ (OQ -2) * 8 - 8 OQ * 1 ] /(OQ - 2)² + 1
= 1 - 16 /(OQ - 2)²
Derivative is 0 when (OQ - 2)² = 16
OQ - 2 = +4 or -4
OQ = 6 or -2. we take only the positive value.
Minimum value of OP + OQ = 8 * 6/(6-2) + 6
= 12 + 6 = 18
let point A be (8, 2).
equation of L : y = - m x + c. It passes through A.
2 = - m * 8 + c => c = 2 + 8 m
OP = x intercept , ie., value of x when y = 0.
0 = - m * OP + c = - m * OP + 2 + 8 m
=> OP = 2 / m + 8
OQ = y intercept , ie., value of y when x = 0
OQ = c = 2 + 8 m
OP + OQ = 8 m + 10 + 2 / m
derivative of (OP + OQ) wrt m : 8 - 2 / m²
derivative = 0 when: m = +1/2 or -1/2. we take only the positive value.
minimum value of OP + OQ = 4 + 10 + 4 = 18
=========================================
simpler method:
The equation of L in the intercept form: x/OP + y/OQ = 1
We are given that OP and OQ are positive. As L passes through A (8, 2):
8/OP + 2/OQ = 1
8 OQ + 2 OP = OP * OQ
OP = 8 OQ / (OQ - 2)
The sum of intercepts : OP + OQ = 8 OQ / (OQ - 2) + OQ
Derivative of OP + OQ wrt OQ: [ (OQ -2) * 8 - 8 OQ * 1 ] /(OQ - 2)² + 1
= 1 - 16 /(OQ - 2)²
Derivative is 0 when (OQ - 2)² = 16
OQ - 2 = +4 or -4
OQ = 6 or -2. we take only the positive value.
Minimum value of OP + OQ = 8 * 6/(6-2) + 6
= 12 + 6 = 18
Answered by
4
Answer:
Step-by-step explanation:
The equation of L in the intercept form: x/OP + y/OQ = 1
We are given that OP and OQ are positive. As L passes through A (8, 2):
8/OP + 2/OQ = 1
8 OQ + 2 OP = OP * OQ
OP = 8 OQ / (OQ - 2)
The sum of intercepts : OP + OQ = 8 OQ / (OQ - 2) + OQ
Derivative of OP + OQ wrt OQ: [ (OQ -2) * 8 - 8 OQ * 1 ] /(OQ - 2)² + 1
= 1 - 16 /(OQ - 2)²
Derivative is 0 when (OQ - 2)² = 16
OQ - 2 = +4 or -4
OQ = 6 or -2. we take only the positive value.
Minimum value of OP + OQ = 8 * 6/(6-2) + 6
= 12 + 6 = 18
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