1. Stating the problem: We have an acute triangle with the longest side measuring 30 inches and the other two sides congruent but unknown in length. 2. Let the length of each congruent side be $x$. 3. Since the triangle is acute, the square of the longest side must be less than the sum of the squares of the other two sides: $$30^2 < x^2 + x^2$$ 4. Simplify the inequality: $$900 < 2x^2$$ 5. Divide both sides by 2: $$450 < x^2$$ 6. Taking the square root of both sides: $$x > \sqrt{450} = \sqrt{225 \times 2} = 15\sqrt{2} \approx 21.213$$ 7. The perimeter $P$ is: $$P = 30 + 2x$$ 8. To minimize the perimeter, choose the smallest $x$ satisfying the inequality: $$x \approx 21.213$$ 9. Calculate the smallest perimeter: $$P \approx 30 + 2 \times 21.213 = 30 + 42.426 = 72.426$$ 10. Round to the nearest tenth: $$72.4 \text{ inches}$$. Final answer: The smallest possible perimeter rounded to the nearest tenth is 72.4 inches.