What is the maximum number of x-intercepts that a polynomial of degree 7 can have
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Answer:
<em>Maximum value of f=2.41</em>
Step-by-step explanation:
<u>Lagrange Multipliers</u>
It's a method to optimize (maximize or minimize) functions of more than one variable subject to equality restrictions.
Given a function of three variables f(x,y,z) and a restriction in the form of an equality g(x,y,z)=0, then we are interested in finding the values of x,y,z where both gradients are parallel, i.e.
for some scalar called the Lagrange multiplier.
For more than one restriction, say g(x,y,z)=0 and h(x,y,z)=0, the Lagrange condition is
The gradient of f is
Considering each variable as independent we have three equations right from the Lagrange condition, plus one for each restriction, to form a 5x5 system of equations in .
We have
Let's compute the partial derivatives
The Lagrange condition leads to
Operating and simplifying
Replacing the value of in the two first equations, we get
From the first equation
Replacing into the second
Or, equivalently
Squaring
To solve, we use the restriction h
Multiplying by 100
Replacing the above condition
Solving for x
We compute the values of y by solving
For
And for
Finally, we get z using the other restriction
Or:
The first solution yields to
And the second solution gives us
Complete first solution:
Replacing into f, we get
f(x,y,z)=-0.4
Complete second solution:
Replacing into f, we get
f(x,y,z)=2.4
The second solution maximizes f to 2.4