Name: maximizing bakery revenue with calculus
Uploaded: 2026-01-12T12:24:25.000Z
Duration: 1 min 29 s
Channel: JoVE

A store wants to find the best price for a Smart TV to maximize weekly sales revenue. Revenue is the total sales income, calculated as price multiplied by units sold.Finding the best price starts with a demand equation that shows how price and units sold relate. Two key pieces of information define this relationship. First, sales records show that 200 units are sold when the price is 350 dollars.Second, a market study shows a linear relationship between price and units sold, with a slope of minus one-half.Using the point-slope form with this slope and the data point gives the demand equation. Rearranging the equation expresses units sold in terms of price.Substituting this into the revenue formula gives a single-variable quadratic function for revenue.As price increases, revenue first rises, reaches a maximum, and then falls. The highest point on this curve shows the price that gives maximum revenue.This point is found by differentiating the revenue function and setting the derivative to zero.Solving this gives the best price as 225 dollars.This example shows how calculus finds maximum values in real situations.

application of differentiation to business

Application of Differentiation to Business

Calculus offers essential techniques for businesses seeking to optimize pricing strategies and revenue. In this case, a bakery wants to determine the ideal price and daily sales volume to maximize revenue. By modeling how changes in price affect demand and revenue, the bakery can apply calculus to make data-driven decisions.The demand function relates the price per cupcake to the number of cupcakes sold and captures how lower prices increase sales. Based on market data, the demand function can be represented as:<figure class="image image-style-align-center" style="display:table;margin-left:auto;margin-right:auto;"><img style="aspect-ratio:300/48;" src="https://cloudfront.jove.com/files/ftp_upload/19838/19838_PT_Equation_1_editor_19838.svg" alt="Polynomial equation $p(x) = 4.6667 - \frac{1}{60}x$, mathematical expression." data-altupdatedat="1775223722000" data-alt="19838_PT_Equation_1.svg" /></figure>Where p(x) is the price per cupcake when x cupcakes are sold per dayThe revenue function, which calculates total revenue, is represented as:<figure class="image image-style-align-center" style="display:table;margin-left:auto;margin-right:auto;"><img style="aspect-ratio:300/48;" src="https://cloudfront.jove.com/files/ftp_upload/19838/19838_PT_Equation_2_editor_19838.svg" alt="R(x)=4.6667x-(1/60)x&#178;; quadratic equation; mathematical analysis; educational use.
" data-altupdatedat="1775223722000" data-alt="19838_PT_Equation_2.svg" /></figure>To maximize revenue, the bakery obtains the marginal revenue function, the derivative of the revenue function:<figure class="image image-style-align-center" style="display:table;margin-left:auto;margin-right:auto;"><img style="aspect-ratio:300/48;" src="https://cloudfront.jove.com/files/ftp_upload/19838/19838_PT_Equation_3_editor_19838.svg" alt="Mathematical derivative formula, R'(x)=4.6667-(1/30)x, educational concept." data-altupdatedat="1775223722000" data-alt="19838_PT_Equation_3.svg" /></figure>By solving for when marginal revenue equals zero, the bakery finds the optimal sales level of:<figure class="image image-style-align-center image_resized" style="display:table;margin-left:auto;margin-right:auto;width:50%;"><img style="aspect-ratio:300/24;" src="https://cloudfront.jove.com/files/ftp_upload/19838/19838_PT_Equation_4_editor_19838.svg" alt="x=140 cupcakes per day, equation for rate calculation, diagram of supply chain analysis." data-altupdatedat="1775223722000" data-alt="19838_PT_Equation_4.svg" /></figure>At this sales level, the price is approximately $2.33 per cupcake.This example demonstrates how calculus links pricing, demand, and revenue into a cohesive optimization process, enabling businesses to make informed decisions that maximize financial outcomes using mathematical modeling.

Calculus offers essential techniques for businesses seeking to optimize pricing strategies and revenue. In this case, a bakery wants to determine the ...

application-of-differentiation-to-business

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JoVE Core

Calculus

Unit

Applications of Differentiation

KEY TERMS AND CONCEPTS

maximizing bakery revenue with calculus

Calculus helps businesses make pricing decisions that can improve revenue. In this example, a bakery uses a cupcake model to find the best price and daily sales volume. The goal is to see how changes in price affect demand and total revenue.

The demand function shows the link between price and how many cupcakes are sold each day. In this model, p(x) gives the price per cupcake when x cupcakes are sold. Lower prices lead to higher demand, so the function captures how sales change with price.

The revenue function gives the bakery’s total income from cupcake sales. To find the point where revenue is highest, the bakery takes the derivative of the revenue function. This derivative is called marginal revenue, which shows how revenue changes when sales increase by one more cupcake.

The bakery then solves for the point where marginal revenue equals zero. That sales level is 150 cupcakes per day. At that point, the price is about $2.33 per cupcake. This example shows how calculus connects price, demand, and revenue in a real business model.

Maximizing Bakery Revenue with Calculus

absolute and local extreme values

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critical numbers and the closed interval method

critical-numbers-and-the-closed-interval-method

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rolle’s theorem

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first-derivatives-and-the-shape-of-a-graph

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second derivatives and the shape of a graph

second-derivatives-and-the-shape-of-a-graph

Using Second Derivatives to Read Curvature

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second derivative test: problem solving

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guidelines for sketching a curve

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slant asymptotes

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optimization problems

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Calculus helps businesses make pricing decisions that can improve revenue. In this example, a bakery uses a cupcake model to find the best price and daily sales volume. The goal is to see how changes in price affect demand and total revenue. The demand function shows the link between price and how many cupcakes are sold each day. In this model, p(x) gives the price per cupcake when x cupcakes are sold. Lower prices lead to higher demand, so the function captures how sales change with price. <br...

newton’s method

Newton’s Method for Finding Roots

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the antiderivative of a function

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graphing antiderivatives

Antiderivative Graph Shape and Slope

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application of antiderivatives: linear motion

application-of-antiderivatives-linear-motion

Finding Car Stopping Distance with Integration

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key terms and concepts

Use calculus to model cupcake price, demand, and revenue. Learn how marginal revenue helps a bakery find the sales level that maximizes income.

Maximizing Bakery Revenue with Calculus | JoVE K12

The JoVE Core Calculus collection provides a comprehensive visual guide to the entire calculus sequence, seamlessly connecting mathematical theory to real-world applications in physics and economics. Through dynamic animated lessons, the course transitions smoothly from foundational single-variable concepts to advanced multivariable topics. Students will explore everything from standard differentiation and integration to spatial vectors, multi-dimensional derivatives, and complex vector fields. By breaking down abstract coordinate systems, infinite series, and spatial motion, this structured approach builds the robust analytical skills and advanced mastery essential for STEM success.

calculus

Maximizing Bakery Revenue with Calculus

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