this is section 3.8 problem 14:\na homeowner plans to enclose a 200 square foot rectangular playground in…

this is section 3.8 problem 14:\na homeowner plans to enclose a 200 square foot rectangular playground in his garden, with one side along the boundary of his property. his neighbor will pay for one third of the cost of materials on that side. find the dimensions of the playground that will minimize the homeowners total cost for materials. follow the steps:\n(a) let the width to be y and the length (the side along the boundary of his property) to be x, and assume that the material costs $1 per foot. then the quantity to be minimized is (expressed as a function of both x and y) c=\n. (use fraction for coefficients.)\n(b) the condition that x and y must satisfy is y=\n.\n(c) using the condition to replace y by x in c, c can then be expressed as a function of x: c(x)=\n.\n(d) the domain of c is (\n,\n). (use infty for ∞.)\n(e) the only critical number of c in the domain is x=\n. (keep 1 decimal place (rounded)). we use the second-derivative test to classify the critical number as a relative maximum or minimum, or neither:\nat the critical number x=\n, the second derivative c(\n) is ---select--- < >. therefore at x=\n, ---select--- < >.\n(f) finally, plug x=\n into the condition of x and y we obtain y=\n.\ntherefore the length and width of the playground that will minimize the homeowners total cost for materials are x=\n feet and y=\n feet, with the side along the boundary of his property equals\n feet.\n

this is section 3.8 problem 14:\na homeowner plans to enclose a 200 square foot rectangular playground in his garden, with one side along the boundary of his property. his neighbor will pay for one third of the cost of materials on that side. find the dimensions of the playground that will minimize the homeowners total cost for materials. follow the steps:\n(a) let the width to be y and the length (the side along the boundary of his property) to be x, and assume that the material costs $1 per foot. then the quantity to be minimized is (expressed as a function of both x and y) c=\n. (use fraction for coefficients.)\n(b) the condition that x and y must satisfy is y=\n.\n(c) using the condition to replace y by x in c, c can then be expressed as a function of x: c(x)=\n.\n(d) the domain of c is (\n,\n). (use infty for ∞.)\n(e) the only critical number of c in the domain is x=\n. (keep 1 decimal place (rounded)). we use the second-derivative test to classify the critical number as a relative maximum or minimum, or neither:\nat the critical number x=\n, the second derivative c(\n) is ---select--- < >. therefore at x=\n, ---select--- < >.\n(f) finally, plug x=\n into the condition of x and y we obtain y=\n.\ntherefore the length and width of the playground that will minimize the homeowners total cost for materials are x=\n feet and y=\n feet, with the side along the boundary of his property equals\n feet.\n

Answer

Explanation:

Step1: Find the relationship between (x) and (y)

Given the area (A = xy=200), so (y=\frac{200}{x}). The cost function (C) (since one - side cost is shared with the neighbor). Let the material cost be ($1) per foot. The cost function (C(x)=x + 2y). Substitute (y=\frac{200}{x}) into (C(x)), we get (C(x)=x + 2\times\frac{200}{x}=x+\frac{400}{x})

Step2: Find the domain of (C(x))

Since (x>0) (length cannot be non - positive), the domain of (C) is ((0,\infty))

Step3: Find the critical numbers

Differentiate (C(x)) with respect to (x). Using the power rule ((x^n)^\prime=nx^{n - 1}), (C^\prime(x)=1-\frac{400}{x^{2}}) Set (C^\prime(x) = 0), then (1-\frac{400}{x^{2}}=0) [ \begin{align*} \frac{x^{2}-400}{x^{2}}&=0\ x^{2}-400&=0\ x^{2}&=400\ x& = 20 \end{align*} ]

Step4: Second - derivative test

Differentiate (C^\prime(x)) to get (C^{\prime\prime}(x)=\frac{800}{x^{3}}) When (x = 20), (C^{\prime\prime}(20)=\frac{800}{20^{3}}=\frac{800}{8000}=0.1>0) Since (C^{\prime\prime}(20)>0), the function (C(x)) has a local minimum at (x = 20)

Step5: Find (y)

When (x = 20), (y=\frac{200}{20}=10)

Answer:

(a) (C=x+\frac{400}{x}) (b) (y = \frac{200}{x}) (d) Domain: ((0,\infty)), Critical number (x = 20) (e) (C^{\prime\prime}(20)=0.1>0), so (C(x)) has a local minimum at (x = 20) (f) (x = 20) feet, (y = 10) feet