Optimization Problems - Free Printable

- The problem asks for the dimensions of a triangle inscribed in a semicircle of radius 10 cm (diameter 20 cm) with one side along the diameter, such that the triangle has maximum area.
- The area A of the triangle is given by $A = \frac{1}{2} b h$, where b is the base (along the diameter) and h is the height.
- Since the triangle is inscribed in a semicircle with diameter 20 cm, the vertex opposite the diameter lies on the semicircle. By the Pythagorean theorem, if we let the height be h, then half the base is $\sqrt{10^2 - h^2} = \sqrt{100 - h^2}$, so the full base $b = 2\sqrt{100 - h^2}$.
- However, the provided equations use a different variable setup: they define $b$ as half the base, so the full base is $2b$. The constraint is $b^2 + h^2 = 10^2 = 100$, but the image shows $b^2 + h^2 = 400$, which implies the radius is 20 cm, not 10 cm. To be consistent with the image's math, we proceed with radius 20 cm (diameter 40 cm), even though the text says 10 cm. This appears to be an error in the problem statement or diagram labeling.
- Assuming the diagram and equations are correct (radius 20 cm, diameter 40 cm), the constraint is $b^2 + h^2 = 400$, so $b = \sqrt{400 - h^2}$.
- The area is $A = \frac{1}{2} b h = \frac{1}{2} \sqrt{400 - h^2} \cdot h = \frac{1}{2} (400 - h^2)^{1/2} h$.
- To maximize A, take the derivative with respect to h:
$\frac{dA}{dh} = \frac{1}{2} \left[ (400 - h^2)^{1/2} \cdot 1 + h \cdot \frac{1}{2} (400 - h^2)^{-1/2} \cdot (-2h) \right]$
$= \frac{1}{2} \left[ (400 - h^2)^{1/2} - \frac{h^2}{(400 - h^2)^{1/2}} \right]$
$= \frac{1}{2} \cdot \frac{(400 - h^2) - h^2}{(400 - h^2)^{1/2}} = \frac{1}{2} \cdot \frac{400 - 2h^2}{(400 - h^2)^{1/2}}$.
- Set $\frac{dA}{dh} = 0$: $400 - 2h^2 = 0 \Rightarrow h^2 = 200 \Rightarrow h = \sqrt{200} = 10\sqrt{2}$ cm.
- Then $b = \sqrt{400 - h^2} = \sqrt{400 - 200} = \sqrt{200} = 10\sqrt{2}$ cm.
- Since b is half the base, the full base is $2b = 20\sqrt{2}$ cm.
- The sides of the triangle are: base = $20\sqrt{2}$ cm, and the two equal legs (since it's isosceles) are each $\sqrt{b^2 + h^2} = \sqrt{200 + 200} = \sqrt{400} = 20$ cm.
- Therefore, the dimensions of the triangle with maximum area are base $20\sqrt{2}$ cm and equal sides 20 cm each.

Note: If the radius is indeed 10 cm (as stated in the text), then the constraint should be $b^2 + h^2 = 100$, leading to $h = \sqrt{50} = 5\sqrt{2}$ cm, half-base $b = 5\sqrt{2}$ cm, full base $10\sqrt{2}$ cm, and equal sides 10 cm. The image's math assumes radius 20 cm. Given the discrepancy, the solution based on the image's equations (radius 20 cm) is presented above.