Find maximum area of a rectangle inscribed in the region bounded by graph of y?
Hi, this is a calculus optimization problem. I'm stuck. Find the maximum area of a rectangle inscribed in the region bounded by the graph of y = (4-x)/(2+x) and the axes. I tried solving this by setting Area = x*((4-x)/(2+x)) and then taking the derivative, but to no avail. What am I missing here? Thank you in advance.
A rectangle is inscribed in a region bounded by the x axis, y axis and graph of?
A rectangle is inscribed in a region bounded by the x axis, y axis and the graph of x+2y-8=0. Approximate the dimensions of the rectangle that will produce the maximum area.
Find the maximum area of a rectangle inscribed in the region bounded by the graph of y = (3 − x)/(4 + x)...?
A rectangle bounded by parabola and the axes is either if 1st or 2nd quadrant: http://imgur.com/a/opnpU Rectangle in 1st quadrant has width = x (0 < x < 3) and height f(x). Rectangle in 2nd quadrant has width = −x (−4 < x < 0) and height f(x). From diagram above, we can see that for any rectangle in 1st quadrant with height h, there is a larger rectangle in 2nd quadrant with the same height. So rectangle with maximum area will be in quadrant 2. A = −x * f(x) ........... −4 < x < 0 A = −x(3−x)(4+x) A = −x(−x²−x+12) A = x³ + x² − 12x A' = 3x² + 2x − 12 = 0 x = (−1±√37)/3 Since −4 < x < 0, then x = (−1−√37)/3 Maximum area = −x(3−x)(4+x) = ((1+√37)/3) ((10+√37)/3) ((11−√37)/3) = 2(55+37√37)/27 ≈ 20.745349157
A rectangle is inscribed in the region bounded by one arch of the graph of y=cosx and the x-axis. What value ?
the area function is 2x cos x , x in [0,π/2]...maximize to find x ≈0.86 with area ≈1.122..note : to find the proper x I used Newton's root finding algorithm for f(x) = 2 cos x - 2x sin x = A '(x)
What is the maximum area of a rectangle that can be inscribed in a parabola x^2 = -2y + 12 if its single side lies on the x axis?
The plot will look something like this :Now, since you want one side of rectangle as [math]x[/math]-axis we can conclude that the rectangle will be above [math]x[/math]-axis (i.e. in Quadrant 1 and 2 ).Now let us assume that one vertex of rectangle is [math](a,0)[/math] . So one side of rectangle i.e side on [math]x[/math]-axis will have length of [math]2a[/math] (Since parabola is symmetric about [math]y[/math]-axis hence the rectangle's side on [math]x[/math]-axis will be symmetric to [math]y[/math]-axis)Now the point on Parabola corresponding to [math](a,0)[/math] will be [math]6-\frac{a^2}{2}[/math]Hence area of Rectangle will be [math]A(a) = (2a) \cdot ( 6 - \frac{a^2}{2}) = 12a - a^3[/math]So to get maximum area we will find the maxima of [math]A(a)[/math] hence differentiating this function we get [math]\frac{dA}{da}= 12 - 3a^2[/math]Now we set this equal to [math]0[/math][math]12 - 3a^2 = 0[/math]so [math]a = \pm 2[/math]Now we can take the second order derivative to check the value at [math]a = 2[/math]. This is less than [math]0[/math] hence it is the maxima.Second order derivative of area function [math]= -6a[/math] which at [math]a = 2[/math] =[math] -12[/math]Now the maximum area of rectangle will be [math]16[/math]
Find the area of the largest rectangle which can be inscribed in the region bounded by the x axis and...?
Hey, I have this problem in my calculus book that i am having a lot of trouble with: Find the area of the largest rectangle which can be inscribed in the region bounded by the x axis and the graph of y = 12 - x^2. (Begin by drawing the rectangle. Let be the distance from the origin to the lower right hand corner of the rectangle. Express the area of the rectangle in terms of X. I am kind of assuming that the rectangle may be a square, but I that is just speculation. It looks like this rectangle is inside a parabola which opens upward. Any information would be much appreciated. Thanks.
Find the height of the largest rectangle that can be inscribed in the region?
Length of the rectangle = 2x Height of the rectangle = y Area = xy A = x √(49 - x^2) Differentiate A with respect to x dA/dx = √(49 - x^2) -2 x^2 / 2 √(49 - x^2) dA/dx = √(49 - x^2) - x^2 / √(49 - x^2) =0 Multiply by √(49 - x^2) (49-x^2)-x^2=0 49=2x^2 x^2=24.5 x= √24.5 =4.9497 y= √(49 - 24.5) =√24.5 =4.9497 The maximum height is √24.5 Verify that d^2A/dx^2 < 0