Can someone help me with these questions?
1) A rectangle has 2 of its vertices on the x axis and the other 2 lie on the parabola y=16-x^2. Find the dimensions of the rectangle if it is to have maximum area
2) Find the dimensions of the rectangle with largest area which can be cut from a circle with equation x^2+y^2=4
Thanks
1) To simplify things, let's just work with one side of the parabola, because then we can just double the width because the parabola is symmetrical.
So we want the blue area to be a maximum and from the diagram we can see the width of the rectangle is 'a' and the height of the rectangle is 'f(a)':
 \implies \text{Area}=a \times (16-a^2)=16a-a^3)
Take the derivative and let it equal 0, so we can find the value of 'a' such that the area is a maximum:
})
Find f(a) - the height of the rectangle:
=16-(\frac{4\sqrt{3}}{3})^2=\frac{32}{3})
Hence the maximum area of the rectangle occurs when the width is
and the height is
(Make sure you double the width value to take into account the other side of the parabola!)
2)
The graph
is a circle with radius 2. Through intuition, the largest area will occur when there is a square placed inside the circle with vertices that touch the circle. Hence the diagonal of the square will be equal to the diameter 4.
Spoiler
Using pythagoras:
Hence the dimensions for a maximum area are a rectangle with width and height equal to
.
You can prove that the largest area will occur when there is a square in the circle by rearranging the equation for the circle for 'y', then applying the same steps I applied to Question 1.
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