Course:MATH102/Question Challenge/1997 December Q4

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Question

What is the area of the largest rectangle that can be inscribed, as shown in the figure, in a right triangle whose sides have lengths 6, 8, 10.

Rectangle inscribed in triangle.

Hints

Hint
The rectangle creates a similar triangle. Use that information to eliminate one of the variables.

Solutions

Solution

The goal is to maximize the area of the rectangle. We label the rectangle width $x$ and the height $y$. So we want to maximize $A=xy$. The problem theoretically could be solved by creating a equation for two hypoteneuses of the two smaller right triangles, but this leads to a messy calculation. In order to avoid most of the algebra we will instead use similar triangles. We use the large and small triangles shown in the diagram. We have labeled the width $x$ and the height $y$ of the rectangle. Sides of the two similar triangles are shown in red and green. ${\frac {8}{6}}={\frac {8-y}{x}}$ ${\frac {4}{3}}x=8-y$ Isolating for $y$: $y=8-{\frac {4}{3}}x$ $y={\frac {24-4x}{3}}$ We use this constraint to eliminate $y$ from the formula for the area of the rectangle: $A=xy=x{\frac {24-4x}{3}}={\frac {1}{3}}(24x-4x^{2})$. Now look for critical points by setting $A'(x)=0$. $A'(x)={\frac {1}{3}}(24-8x)=0$ $0=24-8x$ $8x=24$ $x=3$ We can also find the height from $y={\frac {24-4x}{3}}={\frac {24-4\cdot 3}{3}}={\frac {12}{3}}=4$ Finally, the actual area of the rectangle is $A=xy=3\cdot 4=12$ We know that this will produce the rectangle of maximum area because the other two critical points are global minimums. They are $x=0,y=8$ and $x=6,y=9$. They both produce an area of zero. Alternately, we can use the first derivative test. We found that the first derivative is $A'(x)={\frac {24-8x}{3}}$ We see that this is positive for values of $x$ smaller than 3 and negative for values of $x$ larger than 3 . This tells us that the function is increasing towards the critical point, and then decreasing past it, indicating a local maximum occurs at $x=3$.