Solutions des TPs
649
B- Un second probl` eme d’optimisation
8.1- Pour simplifier les choses, nous ne reproduisons pas la figure tout ` a fait
` a l’´ echelle.
plot.new()
par(xpd=NA)
rect(0,0,1,1)
points(0,1,pch=16)
text(0,1,"A",adj=c(2,-0.1),col="blue")
points(1,1,pch=16)
text(1,1,"B",adj=c(-1,0),col="blue")
points(1,0,pch=16)
text(1,0,"C",adj=c(-0.5,1),col="blue")
points(0,0,pch=16)
text(0,0,"D",adj=c(1.7,1),col="blue")
points(0.5,0,pch=16)
text(0.5,0,"H",adj=c(1.5,-0.5),col="blue")
points(1,0.4,pch=16)
text(1,0.4,"Q",pos=4,col="blue")
points(0.5,0.4,pch=16)
text(0.5,0.4,"M",adj=c(0.5,-1),col="blue")
segments(0.5,0,0.5,0.4)
segments(0.5,0.4,1,0.4)
polygon(x=c(0.5,0,1),y=c(0.4,1,1),col=rgb(t(col2rgb("brown"))/256,
alpha=20/100),border="brown")
x <- seq(from=0.6,to=0.565,length=100)
points(x,sqrt(0.1^2-(x-0.5)^2)+0.4,type="l")
text(0.61,0.45,expression(alpha))
8.2- On a cos(α) = MQ/MB donc MB = 5/ cos(α) et MA = MB. MH = BC − BQ,
tan(α) = BQ/MQ = BQ/(AB/2) donc BQ = 5 tan(α) et MH = 6 − 5 tan(α).
D’o` u g(α) = 10/ cos(α) + 6 − 5 tan(α) = (10 − 5 sin(α))/ cos(α) + 6 = 5(2 −
sin(α))/ cos(α) + 6.
8.3- g <- function(alpha) {5 * (2-sin(alpha))/cos(alpha)+6}
8.4- optimize(g,lower=0,upper=pi/2,tol=0.000001)
8.5- g
(α) = 5(− cos
2 (α) − (2 − sin(α))(− sin(α)))/ cos
2 (α) = 5(− cos
2 (α) + 2 sin(α) −
sin
2 (α))/ cos
2 (α) = 5(2 sin(α) − 1))/ cos
2 (α) car cos
2 (α) + sin
2 (α) = 1.
8.6- D(expression(5 * (2-sin(alpha))/cos(alpha)+6),"alpha")
8.7- gprime <- function(alpha) {5 * (2 * sin(alpha)-1)/(cos(alpha))^2}
8.8- uniroot(gprime,lower=0,upper=pi/2,tol=0.000001)$root