A rectangle is constructed with it's base on the x-axis and the two of its vertices on the parabola #y=49 - x^2#. What are the dimensions of the rectangle with the maximum area?

Question

Charles Arocha · Accepted Answer

In other words, we're constructing a rectangle under a dome-shaped form.

The parabola is a 'mountain'-type (because the coefficient of #x^2# is negative. Also, it is symmetrical in respect to the #y#-axis, because there is no #x#-term. We can now simplify the problem to finding a rectangle with vertices at #(0,0) , (x,0), (0,y) and (x,y)# and then double the #x#-values. The area will then be #A=x*y# If we substitute the equation of the parabola for #y#: #A=x*(49-x^2)=49x-x^3# To find the extremes (max of min) we need the derivative and set it to zero: #A'=49-3x^2=0->x^2=49/3->x=sqrt(49/3)~~4.04...# (remember we will have to double that) Use this in the original function: #y=49-x^2=49-49/3=98/3~~32.67# Answer : Dimensions will be #8.08" x "32.67# graph{49-x^2 [-65.4, 66.33, -13.54, 52.3]}

Isabella Ackerman · Answer

undefined The dimensions of the rectangle with the maximum area are: base length = 14, height = 98.