A consumer electronics company makes two different types of smart phones, the j8 and the j8+. Suppose that j9 corresponds to a new smart pho

Question

A consumer electronics company makes two different types of smart phones, the j8 and the j8+. Suppose that j9 corresponds to a new smart phone produced by the company. The manufacturing cost includes labor, materials, and overhead (facilities, etc.). The company’s costs (in dollars) per unit for each type are summarized in the following table.
j8 j8+ j9
Labor 57 73 81
Materials 93 101 113
Overhead 29 34 38
a. Suppose T is the linear transformation that takes as input a vector of unit counts for j8’s, j8+’s, and j9’s respectively, and produces for output a vector of total labor, material, and overhead, respectively. Find a formula for T. (A graphing calculator is recommended.)
b. Determine T −1, and use it to find the production level for each type of phone that will result in the given costs.
Labor = $1955, Materials = $2815, Overhead = $931
(j8, j8+, j9) = ???

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Serenity 2 weeks 2022-01-06T06:20:00+00:00 1 Answer 0 views 0

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    2022-01-06T06:21:48+00:00

    Answer:

    A) The Matricial formula for T is:

    T(V)=\left[\begin{array}{ccc}57&73&81\\93&101&113\\29&34&38\end{array}\right] V

    B) The type of phones produced are:

    \left[\begin{array}{ccc}J8\\J8+\\J9\end{array}\right] =\left[\begin{array}{ccc}7\\8\\12\end{array}\right]

    Step-by-step explanation:

    Let us define the transformation where:

    V=(j8, j8+, j9)=(A, B, C)

    T(V)=(T_{Labor}, T_{materials},T_{Overhead})

    In this case:

    T_{Labor}=57A+73B+81C\\T_{materials}=93A+101B+113C\\T_{Overhead}=29A+34B+38C

    If we define a base for this transformation for the space V/:

    V∈V/  and W’ a base of the vectorial space V/:

    W'=\{(1,0,0)^T(0,1,0)^T(0,0,1)^T\}

    T(V)∈W/ and W” a base of the vectorial space W/:

    W''=\{(1,0,0)^T(0,1,0)^T(0,0,1)^T\}

    The transformation T defined in the base W’ is:

    T(w'_1)=T(1,0,0)=(57, 93,29)^T\\T(w'_2)=T(0,1,0)=(73,101,34)^T\\T(w'_3)=T(0,0,1)=(81, 113,38)^T

    Therefore the transformation’s Matrix in the bases W’ and W” is defined as:

    T(V)_{W''}=\left[\begin{array}{ccc}57&73&81\\93&101&113\\29&34&38\end{array}\right]_{W'W''} V_{W'}

    In this case, the bases W’ and W” are the canonical ones. Therefore we can simply write the formula as:

    T(V)=\left[\begin{array}{ccc}57&73&81\\93&101&113\\29&34&38\end{array}\right]V

    To determine T⁻¹ we need to know if there is some element in the nullspace.

    If there is no element in the nullspace, this transformation is an isomorphism, therefore there is a single solution for T⁻¹. To see quickly if there are elements in the nullspace, we can use the determinant of the transformation’s matrix. If this determinant is different than 0, there aren’t any elements in the nullspace.

    Det|\left[\begin{array}{ccc}57&73&81\\93&101&113\\29&34&38\end{array}\right]|=-116\neq 0 \leftrightarrow Null([T]=\{0\}

    we can calculate the inverse matrix of [T] knowing that it exists. In this case:

    [T]^{-1}=\left[\begin{array}{ccc}1/29&5/29&-17/29\\257/116&183/116&-273/29\\-233/116&-179/116&258/29\end{array}\right]

    Therefore the inverse transformation of T is defined as:

    T^{-1}(W)=[T]^{-1}W=[T]^{-1}(T(V_?))

    In B), we know that T(V?) is (1955,2815,931). Therefore if we replace it in the equation above:

    [T]^{-1}((1955,2815,931)^T)=\left[\begin{array}{ccc}1/29&5/29&-17/29\\257/116&183/116&-273/29\\-233/116&-179/116&258/29\end{array}\right] \cdot \left[\begin{array}{ccc}1955\\2815\\931\end{array}\right] =\left[\begin{array}{ccc}7\\8\\12\end{array}\right]

    Therefore the type of phones produced are:

    \left[\begin{array}{ccc}J8\\J8+\\J9\end{array}\right] =\left[\begin{array}{ccc}7\\8\\12\end{array}\right]

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