Find the point p on the line yequals4x that is closest to the point (51 comma 0 ). what is the least distance between p and (51 comma 0 )? let d be the distance between the two points. what is the objective function in terms of one number, x?

Step-by-step explanation:

Use the distance formula D = √[ (x - x0)² + (y - y0)² ], using (x0, y0):  (51, 0).  Replace 'y' with '4x':

D = √[ (x - 51)² + (4x - 0)² ], or (after simplification)

D = √[ x² - 102x + 2601) + 16x² ], or  D = √[ 17x² - 102x + 2601 ].

This is the objective function in terms of one variable (x).

You don't specify the method to be used.  The main method choices you have are (1) calculus and (2) algebra.  

We want the x value at which this distance D is a minimum.  We can find that x value by finding the minimum of 17x² - 102x + 2601), whose graph is that of a parabola with minimum at x = -b/(2a).  Here, that x is x = 102/(34), or x = 3.  

We conclude that the distance between the given point and the given line is a minimum when x = 3.

That distance is D = √[17x² - 102x + 2601], evaluated at x = 3:

√[ 17(9) - 102(3) + 2601 ] = 49.5 approximately

13.33%

say 50 oranges, then 100 apples

80 are red, 20 are yellow

total fruit = 150

20chances out of 150 for yellow

20/150 = 13.33%