![\"\"](\"/images/stars/stars-10.gif\")
You can put this solution on YOUR website!
(1) The area is 20, so if we say the sides are length x and y, we can say xy=20. The perimeter is 2x+2y. Solving for y in the first equation(area) gives y=20/x, so we can substitute this into the second equation(perimeter) to get 2x+2(20/x) =2x+40/x. So this is the function for the perimeter in terms of x.
\n" ); document.write( "You can say P(x)=2x+40/x.\r
\n" ); document.write( "\n" ); document.write( "(2) The maximum value for y is going to occur at the vertex of the parabola. In order to find this point, could write this parabola in the form y-k=a(x-h)^2 by completeing the square. So, y+1=-x^2-5x, so y+1=-(x^2+5x), so y+1-(5/2)^2=-(x^2+5x+(5/2)^2), so y-21/4 = -(x+5/2)^2, so the vertex is (-5/2,21/4), so the maximum value for y is 21/4.\r
\n" ); document.write( "\n" ); document.write( "This is a very straightforward problem if you are in calculus, but I am not sure if you are or not. In calculus, you find the derivative and set that equal to zero to find the maximum/minimum. so the derivative is dy/dx=-2x-5, so solving -2x-5=0 gives x=-5/2. So the maximum y is attained at x=-5/2, so y=-(-5/2)^2-5(-5/2)-1 = -25/4 + 25/2 -1 = 25/4 - 1 = 21/4. \n" ); document.write( "