Question

A rectangle is to be inscribed in a right triangle having sides of length 6 in, 8 in, and 10 in. Find the dimensions of the rectangle with greatest area assuming the rectangle is positioned as in Figure 1. Figure1

Answer 1

Answer:  width = 2.4 in, length = 5

Step-by-step explanation:

The max area of a right triangle is half the area of the original triangle.

Area of the triangle = (6 x 8)/2  = 24

--> area of rectangle = 24 ÷ 2 = 12

Next, let's find the dimensions.

The length is adjacent to the hypotenuse. Since we know the area is half, we should also know that the length will be half of the hypotenuse.

length = 10 ÷ 2 = 5

Use the area formula to find the width:

A = length x width

12 = 5 w

12/5 = w

2.4 = w

Answer 2

The dimensions of the rectangle with greatest area is length is 3 inch and the width is 4 inch.

Let the length and width of the rectangle be x and y.

Then Area of the rectangle = xy

Now, from the triangle we can conclude that

$\frac{6-x}{y} =\frac{6}{8} \\y=8(\frac{6-x}{6} ).$

Put the value of y in Area we get

$A(x)=x\frac{8}{6} (6-x)\\A(x)=\frac{8}{6}(6x-x^{2} )$

Differentiating it w.r.t x we get

$A'(x)=\frac{8}{6}(6-2x )\\A''(x)=\frac{8}{6}(0-2 )\\A''(x)=\frac{-8}{3}$

Put A'(x)=0 for maximum /minimum value

$A'(x)=0\\\frac{8}{6}(6-2x)=0\\x=3$

Now, $A''(3)=-\frac{8}{3} <0$

Therefore the area of the rectangle is maximum for x=3 inch

Now,

$y=\frac{8}{6} (6-3)\\y=4$

Thus the dimensions of the rectangle with greatest area is 3 inch by 4 inch.

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