1. Optimization is done by finding the derivative of a function, then setting it equal to zero. The final step is to solve for the variable. We use the derivative and the process of optimization to find the maximum or minimum. For example,
2. Find the point on the on the line y=2x+3 that is closest to the origin. To optimize, use the distance formula and the original equation. d= sqrt[(x-0)^2 + (y-0)^2]
Plug in the original equation for y in the distance formula. d= sqrt[x^2 + (2x+3)^2]
Remove the square root and take the derivative of the equation. (d^2)' = 2x + 2*(2x+3)*2
Solve for x. (d^2)'= 2x +8x+12
Set the equation equal to zero to solve. 0 = 10x + 12 -12 = 10x x= -12/10 x= -6/5
Plug the x value back into y equals to find the point that is asked for in the original question and solve for y. y = 2(-6/5)+3 y = -12/5 + 3 y = 3/5
So the answer to the point that is closest to the origin is (-6/5, 3/5)
3. If f'(x) = 6x^2 -10x -1 then,
a)f(x) = 2x^3 -5x^2 -x + 7
b)f(x) = 2x^3 -5x^2 -x -15
Dane you are a great friend and even better scholar. This is fantastic work, couldn't have done it better myself. A little more specific on the steps of optimization would be good!
ReplyDeleteWhat a brilliant post. I enjoyed your organization. It was also convenient how you included the explanation with the example instead of separately.
ReplyDeleteWhat about solving for a variable before finding the derivative? This is a common theme for most problems we have done.
ReplyDeleteI saw this while scrolling through Google+ and would like to thank Dane for his exceptional math skills.
ReplyDelete