Very Tricky Calculus Problem~ Optimization

I have a Very Tricky Calculus Optimization Problem which i need help with.... It goes like this:
"A boat on the ocean is "5" miles from the nearest point, "B", on a straight shore line; that point is 8 miles from a restaurant on the shore. A woman plans to reow the boat straight to a point, "P", on the shore and then walk along the shore to the restaurant. If she rows at 3 mi/hr and walks at 4 mi/hr, at which point, "P", on the shore should she land to "minimize" the total travel "Time" (not distance)....?????"
please help me to understand.

Let x be the distance along the shoreline from point B to point P. Then
the distance the woman will row is sqrt(5^2 + x^2), and the distance she
will walk is 8 - x. Since time = distance/speed, we want to minimize

T(x) = sqrt(25 + x^2)/3 + (8 - x)/4. To do this find dT/dx and set equal to 0:

dT/dx = (1/3)*(1/2)*(25 + x^2)^(-1/2) *(2x) - 1/4 = (x/3)*(25 + x^2)^(-1/2) - 1/4

which equals 0 when x = (3/4)*(25 + x^2)^(1/2). Square both sides to get

x^2 = (9/16)*(25 + x^2) ---> (7/16)x^2 = 225/16 ---> x^2 = 225/7 --->

x = 5.67 miles, as x must be positive.

Note that d/dx(dT/dx) = 25/(25 + x^2)^(3/2) > 0 for x = 5.67 so by the second
derivative test T is minimized at x = 5.67 miles.

DRAW the 'picture ' ...O = boat , R = restaurant ,

| BP | = x , | PR | = 8 - x , | OP | = √ ( x² + 25 )...time = time on water + time on land

T = √ ( x² + 25 ) / 3 + [ 8 - x ] / 3...minimize T....this is a D = rt problem