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Chapter 8: Problem 14

A box with no top is to be made from a \(22 \mathrm{~cm}\) by \(28 \mathrm{~cm}\) piece of card board by cutting squares of equal size from each corner and folding up the 'tabs'. What size of squares should be cut from each corner to make the box of largest volume?

Short Answer

Expert verified
Cut squares of approximately 3.41 cm from each corner for maximum volume.

Step by step solution

01

Understand the Problem

We need to create a box with no top from a piece of cardboard that measures \( 22 \text{ cm} \) by \( 28 \text{ cm} \). By cutting squares of side \( x \) cm from each corner and folding the sides up, the goal is to determine \( x \) such that the volume of the box is maximized.
02

Express Volume in Terms of x

Once the squares are cut from each corner, the dimensions of the box will be \( (22-2x) \) cm by \( (28-2x) \) cm with a height of \( x \) cm. The volume \( V \) of the box can be expressed as:\[ V = x(22-2x)(28-2x) = x(616 - 100x + 4x^2) \]
03

Simplify the Volume Equation

Expand the equation:\[ V = 4x^3 - 100x^2 + 616x \]This is the cubic function in terms of \( x \) that represents the volume of the box.
04

Determine Critical Points

To find the maximum volume, take the derivative of the volume and set it equal to zero:\[ V' = 12x^2 - 200x + 616 \]Set \( V' = 0 \) to find critical points:\[ 12x^2 - 200x + 616 = 0 \]
05

Solve the Quadratic Equation

Solve the quadratic equation using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 12 \), \( b = -200 \), and \( c = 616 \).Calculating:\[ x = \frac{200 \pm \sqrt{(-200)^2 - 4 \(12\) \(616\)}}{2 \(12\)} \]\[ x = \frac{200 \pm \sqrt{40000 - 29568}}{24}\]\[ x = \frac{200 \pm \sqrt{10432}}{24} \]\[ x = \frac{200 \pm 102.13}{24} \]\[ x \approx 12.59 \, \text{cm}\] or \[ x \approx 3.41 \, \text{cm}\]
06

Evaluate the Critical Points

Substitute potential values of \( x \) into the volume equation to determine which gives the largest volume. Since \( x = 12.59 \, \text{cm} \) results in negative length or width, only \( x = 3.41 \, \text{cm} \) should be considered. Calculate and verify the volume is maximized at this \( x \) value.
07

Consider the Constraints

Ensure that the choice for \( x \) is within feasible constraints, i.e., \( 0 < x < 11 \) because length or width cannot be negative. Taking \( x = 3.41 \) cm is reasonable within these constraints.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cubic Functions
Cubic functions are polynomial functions of degree three. In this context, our volume equation is a cubic function:
  • It has the general form: \( ax^3 + bx^2 + cx + d \).
  • The equation for the box's volume simplifies to \( V = 4x^3 - 100x^2 + 616x \).
  • This function helps us calculate the box's volume based on the size \( x \).
Cubic functions are very important in calculus for modeling real-world scenarios involving rates of change. They often have turning points where they change direction, which is crucial in optimization problems like this one.
To find the optimal solution, we need to identify these turning points, also known as critical points.
Derivative of a Function
Derivatives are fundamental when attempting to find critical points, helping us understand how the function behaves. The derivative of a function provides insight into how the function's output changes with respect to its input. In our problem, the derivative of the volume function, \( V' = 12x^2 - 200x + 616 \), is key to finding the box size that maximizes volume:
  • Taking the derivative allows us to find slopes of tangent lines to the curve of the volume function.
  • By setting the derivative equal to zero, \( V' = 0 \), we can find possible critical points, where the slope of the tangent is zero.
  • These points indicate where the volume stops increasing or decreasing.
This process is called finding the critical points, and it's a cornerstone of optimization in calculus.
Quadratic Equation
To solve for the critical points, we need to use what's called a quadratic equation. Here, we've got a derivative quadratic equation \( 12x^2 - 200x + 616 = 0 \). This equation is solved using the quadratic formula:
  • \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 12 \), \( b = -200 \), and \( c = 616 \).
  • Plug in the values to find \( x \) and you get approximate values of 12.59 cm and 3.41 cm.
The quadratic formula is a powerful tool in algebra, allowing us to find the roots of any quadratic equation, which represent the potential solutions for maximizing or minimizing a function, in this case, the volume of the box.
Critical Points
Critical points are where the function's derivative is zero, indicating potential maxima or minima. For the box problem, these points represent potential sizes for the square cuts that might maximize the box volume. After solving the quadratic equation, the values of \( x \) found are 12.59 cm and 3.41 cm:
  • However, we must consider practical constraints. The length \( 22 - 2x \) and width \( 28 - 2x \) must remain positive.
  • This restricts feasible \( x \) values to between 0 and 11 cm.
  • Therefore, only \( x = 3.41 \) cm is a practical solution, while 12.59 cm would make dimensions negative.
By checking these constraints, we confirm which critical point is valid, ensuring our solution is both mathematically and practically sound.

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Exercise 8.3.5 Sickle cell anemia is an inherited blood disease in which the body makes sickle-shaped red blood cells. It is caused by a single mutation from glutamic acid to valine at position 6 in the protein Hemoglobin B. The gene for hemoglobin \(\mathrm{B}\) is on human chromosome \(11 ;\) a single nucleotide change in the codon for the glutamic acid, GAG, to GTG causes the change from glutamic acid to valine. The location of a genetic variation is called a locus and the different genetic values (GAG and GTG) at the location are called alleles. People who have GAG on one copy of chromosome 11 and GTG on the other copy are said to be heterozygous and do not have sickle cell anemia and have elevated resistance to malaria over those who have GAG on both copies of chromosome 11 . Those who have GTG on both copies of chromosome 11 are said to be homozygous and have sickle cell anemia \(-\) the hydrophobic valine allows aggregation of hemoglobin molecules within the blood cell causing a sickle- like deformation that does not move easily through blood vessels. Let \(A\) denote presence of \(\mathrm{GAG}\) and \(a\) denote presence of GTG on chromosome \(11,\) and let \(A A,\) \(A a\) and \(a a\) denote the various presences of those codons on the two chromosomes of a person (note: \(A a=a A) ; A A, A a\) and aa label are the genotypes of the person with respect to this locus. It is necessary to assume non-overlapping generations, meaning that all members of the population are simultaneously born, grow to sexual maturity, mate, leave offspring and die. Let \(P, Q\) and \(R\) denote the frequencies of \(A A, A a\) and aa genotypes in a breeding population and let \(p\) and \(q\) denote the frequencies of the alleles \(A\) and \(a\) among the chromosomes in the same population. The frequencies \(P, Q,\) and \(R\) are referred to as genotype frequencies and \(p\) and \(q\) are referred to as allele frequencies. In a population of size, \(N\), there will be \(2 N\) chromosomes and \(P \times 2 N+Q \times N\) of the chromosomes will be \(A\). In a mating of \(A A\) with \(A a\) adults, the chromosome in the fertilized egg (zygote) obtained from \(A A\) must be \(A\) and the chromosome obtained from \(A a\) will be \(A\) with probability \(1 / 2\) and will be \(a\) with probability \(1 / 2\). Therefore, the zygote will be \(A A\) with probability \(1 / 2\) and will be \(A a\) with probability \(1 / 2\) a. Show that the allele frequencies \(p\) and \(q\) in a breeding population with genotype frequencies \(P, Q\) and \(R\) are given by $$ p=P+\frac{1}{2} Q $$ and $$ q=\frac{1}{2} Q+R $$ b. Assume a closed population (no migration) with random mating and no selection. Complete the table showing probabilities of zygote type in the offspring for the various mating possibilities, the frequencies of the mating possibilities, and the zygote genotype frequencies. Include zeros with the zygote type probabilities but omit the zeros in the zygote genotype frequencies. Random mating assumes that the selection of mating partners is independent of the genotypes of the partners. c. When the table is complete, you should see that $$ \begin{aligned} \Sigma_{A a} &=\frac{1}{2} P Q+P R+\frac{1}{2} Q P+\frac{1}{2} Q^{2}+\frac{1}{2} Q R+R P+\frac{1}{2} Q R \\ &=P Q+2 P R+\frac{1}{2} Q^{2}+Q R=2 P\left(\frac{1}{2} Q+R\right)+Q\left(\frac{1}{2} Q+R\right) \\ &=(2 P+Q)\left(\frac{1}{2} Q+R\right) \quad=\quad 2\left(P+\frac{1}{2} Q\right)\left(\frac{1}{2} Q+R\right) \\ &=2 p q \end{aligned} $$ Show that $$ \Sigma_{A A}=p^{2} \quad \text { and } \quad \Sigma_{a a}=q^{2} $$ This means that under the random mating hypothesis, the zygote genotype frequencies of the offspring population are determined by the allele frequencies of the adults. This is referred to as the Hardy-Weinberg theorem. If the probability of an egg growing to adult and contributing to the next generation of eggs is the same for all eggs, independent of genotype, then the allele frequencies, \(p\) and \(q,\) are constant after the first generation. Random mating does not imply the promiscuity that might be imagined. It means that the selection of mating partner is independent of the genotype of the partner. In the United States, blood type would be a random mating locus; seldom does a United States young person inquire about the blood type of an attractive partner. In Japan, however, this seems to be a big deal, to the point that dating services arranging matches also match blood type. The major histocompatibility complex (MHC) of a young person would seem to be fairly neutral; few people even know their MHC type. It has been demonstrated, however, that young women are repulsed by the smell of men of the same MHC type as their own \(^{4}\). d. Show that in a closed random mating population with no selection, if the frequency of \(A\) in the adults in one generation is \(\hat{p},\) then the frequency of \(A\) in adults in the next generation will also be \(\hat{p}\). e. 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