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13 votes
13 votes
Find the dimensions of the rectangle with the largest area; it has two vertices on the x-axis and two vertices above the x-axis on the graph of

y = 5 − x^2.

User Dsolimano
by
3.3k points

1 Answer

20 votes
20 votes

Answer:

(2/3)√15 wide by 10/3 tall, approx 2.582 by 3.333

Explanation:

You want the dimensions of the largest rectangle that fits under the graph of y = 5 -x² above the x-axis.

Dimensions

The graph of y=5-x² is symmetrical about the y-axis, so the width of it will be x -(-x) = 2x and the height will be 5-x².

Area

The area will be the product of the width and height:

A = WH

A = (2x)(5 -x²) = 10x -2x³

Maximum area

The area will be a maximum where the derivative of the area function is zero.

dA/dx = 10 -6x² = 0

x² = 10/6

x = (√15)/3

The corresponding rectangle dimensions are ...

width = 2x = (2/3)√15

height = 5 -x² = 5 -5/3 = 10/3

The rectangle with largest area is (2/3)√15 wide by 10/3 tall.

Find the dimensions of the rectangle with the largest area; it has two vertices on-example-1
User Dzejms
by
3.1k points
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