Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r.

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Wasither1957 · Accepted Answer

Step 1From the figure, by Pythagoras theorem we can writeb2=r2−x2b=2r2−x2→(1)Step 2Area of Triangle =12 (Base)(Height)A=12(b)(r+x)Use Eqn(1) to substitute the value of bA=12(2r2−x2)(r+x)A(x)=(r+x)r2−x2Step 3Differentiate A(x)A′(x)=d[(r+x)r2−x2]dxUse Product RuleA′(x)=r2−x2⋅d[r+x]dx+(r+x)⋅d[r2−x2]dxA′(x)=r2−x2⋅1+(r+x)⋅d[r2−x2]d(r2−x2)×d(r2−x2)dxA′(x)=r2−x2+(r+x)⋅12r2−x2×(−2x)A′(x)=r2−x2−rx+x2r2−x2Step 4Solve for A′(x)=0r2−x2−rx+x2r2−x2=0

Elizabeth Witte · Accepted Answer

Step 1The equation of circle will bex2+y2=r2Step 2Now, area of triangleA=12(2x)(r+y)=x(r+y)A(y)=r2−y2(r+y)triangle derivativeA′(y)=r2−y2+(r+y)⋅−2y2r2−y2=r2−ry−2y2r2−y2equating to zeror2−ry−2y2r2−y2=0r2−ry−2y2=0y=r±9r2−4⇒y=−r or y=r2rejecting, y=−rTherefore, we check to see y=r2 is maximumA′(y)&amp;gt;0 for y&amp;lt;r2 and A′(y)&amp;lt;0 for y&amp;gt;r2∴y=r2 is maximumx2+(r2)2=r2⇒x=32rDimension of triangle with maximum area is base=2x=2⋅32r=3rh=r+y=r+r2=32r

Find the dimensions of the isosceles triangle of largest area that can be inscri

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