What is the largest possible value of y if (x − 3)^2 + (y + 6)^2 = 100?
a= −2
b= 4
c= 5
d= 8
e= 12
I recognised that for for y to be maximized, the value of (x - 3)^2 must be minimized. Since (x - 3)^2 must be positive, the smallest value it can be is 0.
But then I get stuck at this point y^2 + 12y - 55 = 0
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equation with two variables
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To MAXIMIZE the value of x, we must MINIMIZE the value of (x − 3)².amanjena wrote:What is the largest possible value of y if (x − 3)^2 + (y + 6)^2 = 100?
a= −2
b= 4
c= 5
d= 8
e= 12
Since the square of a value cannot be negative, the smallest possible value of (x − 3)² is 0, when x=3.
Substituting x=3 into (x - 3)² + (y + 6)² = 100, we get:
(3 - 3)² + (y + 6)² = 100
0 + (y + 6)² = 100
(y + 6)² = 100
y + 6 = ±10
y = 4 or y = -16.
Thus, the greatest possible value of y is 4.
The correct answer is B.
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Mathsbuddy
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Rearranged:
(y+6)^2 = 100 - (x-3)^2
The minimum value of (x-3)^2 is 0 (when x = 3)
So maximum value of y comes from:
(y+6)^2 = 100
y+6 = 10 (ignoring y + 6 = -10 as this would yield a lower value)
y = 4
ANSWER B
(y+6)^2 = 100 - (x-3)^2
The minimum value of (x-3)^2 is 0 (when x = 3)
So maximum value of y comes from:
(y+6)^2 = 100
y+6 = 10 (ignoring y + 6 = -10 as this would yield a lower value)
y = 4
ANSWER B
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Mathsbuddy
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Hi Amenjena,amanjena wrote:What is the largest possible value of y if (x − 3)^2 + (y + 6)^2 = 100?
a= −2
b= 4
c= 5
d= 8
e= 12
I recognised that for for y to be maximized, the value of (x - 3)^2 must be minimized. Since (x - 3)^2 must be positive, the smallest value it can be is 0.
But then I get stuck at this point y^2 + 12y - 55 = 0
Just take a step backwards. No need to expand the brackets and simplify.
(y+6)^2 = 100 is a simple square
so y+6 = +-10
therefore yMAX = 10 - 6 = 4
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We can also SOLVE FOR y.amanjena wrote:What is the largest possible value of y if (x − 3)² + (y + 6)² = 100?
a= −2
b= 4
c= 5
d= 8
e= 12
We'll also use the fact that, if y² = k (where k is > 0), then y = +√k or y = -√k
(x − 3)² + (y + 6)² = 100
(y + 6)² = 100 - (x − 3)²
So, applying our green rule, we get: y + 6 = +√[100 - (x − 3)²] or y + 6 = -√[100 - (x − 3)²]
Since our goal is to MAXIMIZE the value of y, we can ignore y + 6 = -√[100 - (x − 3)²]
Let's keep going with y + 6 = +√[100 - (x − 3)²]
Subtract 6 from both sides to get: y = √[100 - (x − 3)²] - 6
So, to MAXIMIZE the value of y, we must MAXIMIZE the value of √[100 - (x − 3)²]
To MAXIMIZE the value of √[100 - (x − 3)²], we must MINIMIZE the value of (x − 3)²
(x − 3)² is minimized when x = 3
So, let's see what y equals when x = 3
We already know that y = √[100 - (x − 3)²] - 6
Plug in to get: y = √[100 - (3 − 3)²] - 6
Simplify: y = √[100 - 0²] - 6
Simplify: y = √[100] - 6
Evaluate: y = 10 - 6 = [spoiler]4 = B[/spoiler]
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Brent
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