A rectangle is constructed with its base on the x-axis and two of its vertices on the parabola y=16−x2 What are the dimensions of the rectangle with the maximum area? What is that area?

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A rectangle is constructed with its base on the x-axis and two of its vertices on the parabola y=16−x2  What are the dimensions of the rectangle with the maximum area? What is that area?

Laura Worden · Accepted Answer

Step 1Step 2The problemA=2xy⇒maxgiveny=16−x2So, we need2x(16−x2)=f(x)⇒maxWe havef′(x)=(32x−2x3)′=32−6x2Equating f′(x)=0⇒x=163=43 is the positive solution.Since f′(x) changes its sign from positive to negative when passing through x=43,there is a maximum at that point.Corresponding y isy=16−x2=16−163=323so the dimensions are83×323and the area isA=83×323=25633Dimensions: 83×323, Area 323

Janet Young · Accepted Answer

Step 1A Rectangle with its base on x-axis and two of its vertices on the parabola whose equation is y=16−x2Let A be the area of a rectangle, the expression is given as:A=2xy⇒maxNow, we need to maximize the area of a rectangle that is,f(x)=2x(16−x2)To maximize the area, differentiate the above function with respect to x and equate it with zero,f′(x)=32−6x232−6x2=0x=163=43≈2.31Put the value of x in the equation of parabola and find the value of y,y=16−x2y=16−163y=323≈10.67In the rectangle with the maximum area, the shorter dimension is 2.31 and longer dimension is 10.76Thus, the maximum area of the rectangle is,A=2×2.31×10.76≈49.29sq. units

A rectangle is constructed with its base on the x-axis and two of its vertices o

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