Henry ran five races, each of which was a different positive integer number of
miles. the mean of his distances was 4 miles. what is the maximum possible distance
of his longest race?

10 miles.

Step-by-step explanation:

Let x be the number of miles on Henry's longest race.

We have been given that Henry ran five races, each of which was a different positive integer number of  miles.

We can set an equation for the average of races as:

As distance covered in each race is a different positive integer, so let his first four races be 1, 2, 3, 4.

Now let us substitute the distances of 5 races as:

Let us multiply both sides of our equation by 5.

Let us subtract 10 from both sides of our equation.

Therefore, the maximum possible distance of Henry's longest race is 10 miles.

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!