Find the maximum value of
1 answer:
Answer:
(3, 1)
Step-by-step explanation:
Given the following constraints:
2x + 4y ≤ 10
x + 9y ≤ 12
x ≥ 0
y ≥ 0
We have to first graph the above constraints and secondly, we find the boundary points for the feasible region. This is done by plotting the graphs of 2x + 4y ≤ 10 and x + 9y ≤ 12.
From the graph, the following points satisfy the constraints:
(0,0) , (3,1) , (0,4/3) , (5,0)
Given that:
P = x + 6y
For (0, 0): P = x + 6y = 0 + 6(0) = 0
For (3, 1): P = x + 6y = 3 + 6(1) = 9
For (0, 4/3) P = x + 6y = 0 + 6(4/3) = 8
For (5, 0) P = x + 6y = 5 + 6(0) = 5
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Solve for <em>x</em> when √(<em>x</em> ² - 4) = 1 :
√(<em>x</em> ² - 4) = 1
<em>x</em> ² - 4 = 1
<em>x</em> ² = 5
<em>x</em> = ±√5
We're looking at <em>x </em>≤ 0, so we take the negative square root, <em>x</em> = -√5.
This means <em>f</em> (-√5) = 1, or in terms of the inverse of <em>f</em>, we have <em>f</em> ⁻¹(1) = -√5.
Now apply the inverse function theorem:
If <em>f(a)</em> = <em>b</em>, then (<em>f</em> ⁻¹)'(<em>b</em>) = 1 / <em>f '(a)</em>.
We have
<em>f(x)</em> = √(<em>x</em> ² - 4) → <em>f '(x)</em> = <em>x</em> / √(<em>x</em> ² - 4)
So if <em>a</em> = -√5 and <em>b</em> = 1, we get
(<em>f</em> ⁻¹)'(1) = 1 / <em>f '</em> (-√5)
(<em>f</em> ⁻¹)'(1) = √((-√5)² - 4) / (-√5) = -1/√5
The sign must be negative; see the attached plot, and take note of the negatively-sloped tangent line to the inverse of <em>f</em> at <em>x</em> = 1.
