P>0 means the opening is upward, are you sure that y = px2+qx (p > 0, q<0) is tangent to x+y=5? Obviously, you can only intersect. Don't trust your certificate. So there is something wrong with the topic.

Guess what? I mistook you for the wrong number first.

I changed the topic to y = px 2+qx (p

Answer:

When p =-4/5 and q = 3, the enclosed area is the largest.

The maximum area is 250/3.

The following process

Y = px 2+qx and x axis intersect at (0,0) (-q/p,0).

Arrange these two functions into the following equations.

y=px^2+qx

y=-x+5

form

px^2+(q+ 1)x-5=0

δ=(q+ 1)^2+20p=0

So p =-(q+ 1) 2/20.

S = ∫ (-q/p, 0) (px 2+qx) dx (note that the interval of definite integral is in the first bracket).

=5q^3/(6p^2)

Substitute p =-(q+ 1) 2/20.

s= 1000*q^3/(3(q+ 1)^4)

When ds/dx =1000/3 * (q 2 * (q+1) 3 * (3-q))/(q+1) 8 = 0.

That is, when q=3 (note that both q and p are ≠0), the area is the largest.

P=-4/5.

Smax=250/3

I am a senior one student, so it is not easy to write this answer. Give a point, thank you!

Oh, and QQ 906 188705.

But I don't think you can add it Send it by e-mail, 906 188705@qq.com.