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a rancher has 400 feet of fencing with which to enclose two adjacent rectangular corrals (see figure). what dimensions should be used so that the enclosed area will be a maximum?
a rancher has 400 feet of fencing with which to enclose two adjacent rectangular corrals (see figure). what dimensions should be used so that the enclosed area will be a maximum?
The dimensions that maximize the enclosed area are:Length of one corral: x=100 feetWidth of one corral: y=100 feetLet's denote the dimensions of the rectangular corrals as follows:Length of one corral: x feetWidth of one corral: y feetSince there are two adjacent corrals, the total length of the fencing used will be the sum of the perimeters of the two rectangles. The perimeterP of a rectangle is given by the formula:P=2×(length+width)In this case, the total length of fencing is 400 feet, so:400=2×(x+y)Solving for one of the variables in terms of the other, we get:x+y=200Now, we want to maximize the area A of the enclosed region. The area of one rectangle is given by the formula:A=length×widthSince there are two rectangles, the total area is the sum of the areas of the two rectangles:A=2xyNow, we can express one of the variables in terms of the other using the constraint x+y=200. For example, solving for x:x=200−ySubstitute this expression for x into the area formula:A=2(200−y)ySimplify:A=400y−2y^2Now, we want to find the value of y that maximizes A. To do this, take the derivative of A with respect to y, set it equal to zero, and solve for y:dA/dy =400−4ySetting the derivative equal to zero:400−4y=0Solving for y:y=100Now that we have the value of y, we can find x using the constraintx+y=200:x+100=200x=100So, the dimensions that maximize the enclosed area are:Length of one corral: x=100 feetWidth of one corral: y=100 feet...