As shown in the figure, point P starts from point O and moves in the positive direction of X axis at a speed of one unit per second. When crossing P, the vertical line of the X axis intersects y= 1/2x, and a square PABC is made to the right with PA as one side. When point P moves for 4 seconds, point Q starts from point P and moves along PA-AB-BC at a speed of 2 units per second. When q and c coincide, p?

1) Find the functional relationship between S and T, and write the range of independent variable T?

2) Find the maximum value of s.

Solution:?

Meaning from the topic: When P moves for 4 seconds, the coordinate of P is (4,0); The moving distance of point q is s[Q]=2t, x[P]=(4+t), y[A]=(4+t)/2?

(1), when q is on PA, s[Q]≤y[A]=(4+t)/2, so t≤4/3(s)?

At this time:

S[OPQ]=f(t)=x[P]？ y[Q]/2=(4+t)？ 2t/2=t(t+4)=t^2+4t？

MAX[S[OPQ]]=？ f(4/3)= 16/9+ 16/3 = 64/9？

(2) when q is on AB, y[A]≤s[Q]≤2y[A], so: 4/3≤t≤4(s)?

At this time:

S[OPQ]=？ f(t)=x[Q]？ y[A]/2=(s[Q]-y[A]+x[P])？ *y[A]/2？

=(2t-(4+t)/2+4+t)？ (4+t)/4=(5t+4)？ (t+4)/8=(5t^2+24t+ 16)/8？

MAX[S[OPQ]]=？ f(4)=24×8/8=24？

(3) When q is on BC, 2y[A]≤s[Q]≤3y[A], so: 4≤t≤ 12(s)?

At this time:

S[OPQ]=？ f(t)=x[C]？ y[Q]/2=(x[P]+y[A])？ (3y[A]-s[Q])/2？

=(4+t+(4+t)/2)？ (3(4+t)/2-2t)/2=3(4+t)？ ( 12-t)/8=3(48+8t-t^2)/8？

MAX[S[OPQ]]=？ f(4)=3×8×8/8=24 .？

So the maximum value of s is 24.