1. **State the problem:** We want to find the amount over the cost price that maximizes the revenue for the electronics store. 2. **Define variables:** Let $x$ be the amount over the cost price (in dollars). 3. **Given information:** - At $x=660$, the store sells 52 laptops. - For every $40 increase in $x$, the store sells 2 fewer laptops. 4. **Express the number of laptops sold as a function of $x$:** Since for every $40 increase in $x$, sales decrease by 2 laptops, the slope is $-\frac{2}{40} = -\frac{1}{20}$ laptops per dollar. Let $n(x)$ be the number of laptops sold: $$n(x) = 52 - \frac{1}{20}(x - 660)$$ 5. **Simplify $n(x)$:** $$n(x) = 52 - \frac{x}{20} + \frac{660}{20} = 52 - \frac{x}{20} + 33 = 85 - \frac{x}{20}$$ 6. **Revenue function $R(x)$:** Revenue is price times quantity sold: $$R(x) = x \times n(x) = x \left(85 - \frac{x}{20}\right) = 85x - \frac{x^2}{20}$$ 7. **Maximize revenue:** Take derivative of $R(x)$ with respect to $x$ and set to zero: $$\frac{dR}{dx} = 85 - \frac{2x}{20} = 85 - \frac{x}{10} = 0$$ 8. **Solve for $x$:** $$85 = \frac{x}{10} \implies x = 850$$ 9. **Check second derivative to confirm maximum:** $$\frac{d^2R}{dx^2} = -\frac{1}{10} < 0$$ Since it is negative, $x=850$ maximizes revenue. **Final answer:** The amount over the cost price that maximizes revenue is **850** dollars.